Heat and Gases

Temperature and Thermometers

Temperature Scales

Temperature is a scalar quantity that measures the average kinetic energy of the particles in a substance. Three temperature scales are in common use:

Scale	Symbol	Unit	Absolute Zero	Key Reference Points
Celsius	$T_C$	$^\circ\mathrm{C}$	$-273.15$	$0^\circ\mathrm{C}$ (ice point), $100^\circ\mathrm{C}$ (steam point)
Kelvin	$T$	K	$0$	Same interval as Celsius, shifted by $273.15$
Fahrenheit	$T_F$	$^\circ\mathrm{F}$	$-459.67$	$32^\circ\mathrm{F}$ (ice point), $212^\circ\mathrm{F}$ (steam point)

The Kelvin is the SI unit of temperature and is defined by fixing the Boltzmann constant $k_B = 1.380649 \times 10^{-23}$ J/K.

Conversion Formulae

Celsius to Kelvin and vice versa:

$T = T_C + 273.15$

$T_C = T - 273.15$

Fahrenheit conversions:

$T_F = \frac{9}{5}T_C + 32$

$T_C = \frac{5}{9}(T_F - 32)$

Kelvin to Fahrenheit:

$T_F = \frac{9}{5}(T - 273.15) + 32$

Worked Example 1

Convert $37^\circ\mathrm{C}$ (normal human body temperature) to Kelvin and Fahrenheit.

Solution

$T = 37 + 273.15 = 310.15 \mathrm{ K}$

$T_F = \frac{9}{5}(37) + 32 = 66.6 + 32 = 98.6^\circ\mathrm{F}$

Worked Example 1b

The surface temperature of the Sun is approximately $5778 \mathrm{ K}$. Express this in Celsius.

Solution

$T_C = 5778 - 273.15 = 5504.85^\circ\mathrm{C}$

Thermometric Properties

A thermometer works by exploiting a thermometric property -- a physical property that varies continuously with temperature. Common thermometric properties include:

Thermometer Type	Thermometric Property	Range
Liquid-in-glass	Length of liquid column (thermal expansion)	$-39$ to $350^\circ\mathrm{C}$ (Hg)
Gas thermometer	Pressure at constant volume, or volume at constant pressure	$-270$ to $1500^\circ\mathrm{C}$
Thermocouple	EMF between two junctions at different temperatures	$-200$ to $1500^\circ\mathrm{C}$
Resistance thermometer	Electrical resistance of a metal (e.g., platinum)	$-200$ to $850^\circ\mathrm{C}$

Calibration of a Thermometer

To calibrate an arbitrary thermometer, two fixed points are needed:

1. Lower fixed point (ice point): temperature of pure melting ice at standard pressure ($0^\circ\mathrm{C}$)
2. Upper fixed point (steam point): temperature of pure boiling water at standard pressure ($100^\circ\mathrm{C}$)

If $X_0$ is the thermometric property at $0^\circ\mathrm{C}$ and $X_{100}$ at $100^\circ\mathrm{C}$, then the temperature corresponding to a property value $X$ is:

$\theta = \frac{X - X_0}{X_{100} - X_0} \times 100^\circ\mathrm{C}$

This formula assumes the thermometric property varies linearly with temperature, which is an approximation. Different thermometers will agree exactly at the two fixed points but may differ at intermediate temperatures because their properties do not have the same functional dependence on temperature.

Absolute Temperature Scale

The ideal gas scale defines temperature in terms of the pressure of an ideal gas at constant volume. As pressure approaches zero (low-density limit), all real gases behave ideally, so this scale is independent of the particular gas used.

The absolute zero of temperature ($0$ K) is the theoretical temperature at which all thermal motion ceases. On the Celsius scale this corresponds to $-273.15^\circ\mathrm{C}$.

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Internal Energy and Heat

Internal Energy

The internal energy $U$ of a substance is the total kinetic energy (translational, rotational, vibrational) and potential energy (intermolecular forces) of all its particles.

For an ideal gas, there are no intermolecular forces, so the internal energy is entirely kinetic:

$U = \frac{3}{2}Nk_BT = \frac{3}{2}nRT$

where $N$ is the number of molecules, $n$ is the number of moles, $k_B$ is Boltzmann's constant, and $R$ is the molar gas constant.

Heat

Heat $Q$ is the energy transferred between two systems (or between a system and its surroundings) due to a temperature difference. Heat flows spontaneously from a body at higher temperature to one at lower temperature.

Key distinction: Temperature is a state variable (depends only on the state of the system); heat is a process variable (depends on the path taken between states).

Concept	Symbol	Unit	Nature
Internal energy	$U$	J	State function
Heat	$Q$	J	Process variable
Temperature	$T$	K	State function

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Specific Heat Capacity

Definition

The specific heat capacity $c$ of a substance is the amount of heat required to raise the temperature of $1$ kg of the substance by $1$ K (or $1^\circ\mathrm{C}$, since the intervals are identical).

$c = \frac{Q}{m\Delta T}$

where:

• $Q$ = heat energy supplied (J)
• $m$ = mass (kg)
• $\Delta T$ = temperature change (K or $^\circ\mathrm{C}$)

The SI unit is $\mathrm{J kg}^{-1}\mathrm{ K}^{-1}$.

The molar heat capacity $C$ is the heat required per mole per kelvin:

$C = Mc$

where $M$ is the molar mass.

Common Specific Heat Capacities

Substance	Specific Heat Capacity ($\mathrm{J kg}^{-1}\mathrm{ K}^{-1}$)
Water	$4186$
Ice	$2100$
Aluminium	$900$
Copper	$385$
Iron	$449$
Lead	$128$
Glass	$840$
Air (const. $p$)	$1005$

Water has an exceptionally high specific heat capacity, which is why it is used as a coolant and why coastal climates are more moderate than inland climates.

Heat Capacity of an Object

The heat capacity $C_{\mathrm{obj}}$ of a body is the heat required to raise its temperature by $1$ K:

$C_{\mathrm{obj}} = mc$

Derivation from the First Law

When heat $Q$ is supplied to a body at constant volume (no work done):

$Q = \Delta U = mc\Delta T$

This is valid for solids and liquids where thermal expansion work is negligible. For gases, the specific heat capacity depends on whether the process is at constant volume or constant pressure.

Worked Example 2

A $2.0$ kg copper block at $100^\circ\mathrm{C}$ is placed in $0.5$ kg of water at $20^\circ\mathrm{C}$. Find the final temperature, assuming no heat loss to the surroundings.

Solution

Heat lost by copper $=$ heat gained by water:

$m_c c_c (T_c - T_f) = m_w c_w (T_f - T_w)$

$2.0 \times 385 \times (100 - T_f) = 0.5 \times 4186 \times (T_f - 20)$

$770(100 - T_f) = 2093(T_f - 20)$

$77000 - 770T_f = 2093T_f - 41860$

$77000 + 41860 = 2093T_f + 770T_f$

$118860 = 2863T_f$

$T_f = 41.5^\circ\mathrm{C}$

Worked Example 3

An electric heater rated at $2000$ W heats $3.0$ kg of water from $25^\circ\mathrm{C}$ to $75^\circ\mathrm{C}$. How long does it take, assuming $80\%$ of the energy is absorbed by the water?

Solution

Energy required by the water:

$Q = mc\Delta T = 3.0 \times 4186 \times (75 - 25) = 627900 \mathrm{ J}$

Energy supplied by the heater:

$E = Pt = 2000 \times t$

With $80\%$ efficiency:

$0.80 \times 2000 \times t = 627900$

$t = \frac{627900}{1600} = 392.4 \mathrm{ s} = 6.5 \mathrm{ minutes}$

Method of Mixtures Experiment

Aim: To determine the specific heat capacity of a solid (e.g., a metal block).

Procedure:

1. Measure the mass $m_s$ of the solid.
2. Heat the solid in a water bath to a known temperature $T_s$ (e.g., by boiling water so $T_s \approx 100^\circ\mathrm{C}$).
3. Quickly transfer the solid to an insulated calorimeter containing water of mass $m_w$ at temperature $T_w$.
4. Stir and record the maximum temperature $T_f$.
5. Apply the principle of calorimetry (heat lost $=$ heat gained):

$m_s c_s (T_s - T_f) = (m_w c_w + C_{\mathrm{cal}})(T_f - T_w)$

where $C_{\mathrm{cal}}$ is the heat capacity of the calorimeter.

Sources of error:

• Heat loss to the surroundings during transfer (minimise by working quickly)
• Heat absorbed by the thermometer and stirrer
• Incomplete thermal equilibrium
• Evaporation of water

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Latent Heat

Definition

Latent heat is the energy absorbed or released by a substance during a phase change at constant temperature. The word "latent" means hidden, because this heat does not produce a temperature change.

Quantity	Symbol	Unit	Definition
Specific latent heat of fusion	$l_f$	$\mathrm{J kg}^{-1}$	Heat to melt $1$ kg of solid at its melting point
Specific latent heat of vaporization	$l_v$	$\mathrm{J kg}^{-1}$	Heat to vaporise $1$ kg of liquid at its boiling point

The heat involved in a phase change of mass $m$:

$Q = ml$

where $l$ is the appropriate specific latent heat.

Common Latent Heats

Substance	$l_f$ ($\mathrm{kJ kg}^{-1}$)	$l_v$ ($\mathrm{kJ kg}^{-1}$)	Melting Point ($^\circ\mathrm{C}$)	Boiling Point ($^\circ\mathrm{C}$)
Water	$334$	$2260$	$0$	$100$
Ethanol	$109$	$846$	$-114$	$78$
Aluminium	$397$	$10500$	$660$	$2519$
Copper	$205$	$4730$	$1085$	$2562$
Lead	$23$	$871$	$327$	$1749$

Note that $l_v \gg l_f$ for all substances. Vaporisation requires breaking most intermolecular bonds (particles gain enough energy to escape the liquid), whereas fusion only requires weakening them enough to allow the regular solid structure to break down.

Heating Curve

A heating curve plots temperature against time as a substance is heated at a constant rate:

1. Solid phase: temperature rises steadily (slope depends on $c_{\mathrm{solid}}$).
2. Melting: temperature remains constant at the melting point while latent heat of fusion is absorbed.
3. Liquid phase: temperature rises steadily (slope depends on $c_{\mathrm{liquid}}$).
4. Boiling: temperature remains constant at the boiling point while latent heat of vaporisation is absorbed.
5. Gas phase: temperature rises steadily (slope depends on $c_{\mathrm{gas}}$).

The flat regions on the heating curve correspond to phase transitions where all the supplied heat goes into changing the molecular arrangement rather than increasing kinetic energy.

Cooling Curve

A cooling curve is the reverse: as a substance cools, the temperature drops, plateaus at the condensation point, drops again, plateaus at the freezing point, and then drops in the solid phase.

Supercooling can occur: the liquid may cool below its freezing point before crystallisation begins. When crystallisation starts, the released latent heat causes the temperature to jump back up to the freezing point.

Why Does Temperature Remain Constant During Phase Change?

During melting or boiling, the supplied heat energy is used to overcome the intermolecular forces rather than increase the average kinetic energy of the particles. Since temperature is a measure of average kinetic energy, the temperature remains constant while the potential energy of the system increases.

Worked Example 4

How much energy is required to convert $500$ g of ice at $-20^\circ\mathrm{C}$ to steam at $110^\circ\mathrm{C}$?

Solution

The calculation proceeds in five stages:

Stage 1: Heat ice from $-20^\circ\mathrm{C}$ to $0^\circ\mathrm{C}$:

$Q_1 = mc\Delta T = 0.5 \times 2100 \times 20 = 21000 \mathrm{ J}$

Stage 2: Melt ice at $0^\circ\mathrm{C}$:

$Q_2 = ml_f = 0.5 \times 334000 = 167000 \mathrm{ J}$

Stage 3: Heat water from $0^\circ\mathrm{C}$ to $100^\circ\mathrm{C}$:

$Q_3 = mc\Delta T = 0.5 \times 4186 \times 100 = 209300 \mathrm{ J}$

Stage 4: Vaporise water at $100^\circ\mathrm{C}$:

$Q_4 = ml_v = 0.5 \times 2260000 = 1130000 \mathrm{ J}$

Stage 5: Heat steam from $100^\circ\mathrm{C}$ to $110^\circ\mathrm{C}$ ($c_{\mathrm{steam}} \approx 2010$ $\mathrm{J kg}^{-1}\mathrm{K}^{-1}$):

$Q_5 = mc\Delta T = 0.5 \times 2010 \times 10 = 10050 \mathrm{ J}$

Total energy:

$Q_{\mathrm{total}} = 21000 + 167000 + 209300 + 1130000 + 10050 = 1537350 \mathrm{ J} \approx 1.54 \mathrm{ MJ}$

Worked Example 5

$200$ g of ice at $0^\circ\mathrm{C}$ is added to $400$ g of water at $50^\circ\mathrm{C}$ in an insulated container. Find the final temperature and state of the mixture.

Solution

First, check whether all the ice can melt. The heat available from the water cooling to $0^\circ\mathrm{C}$:

$Q_{\mathrm{available}} = m_w c_w \Delta T = 0.4 \times 4186 \times 50 = 83720 \mathrm{ J}$

Heat required to melt all the ice:

$Q_{\mathrm{melt}} = m_i l_f = 0.2 \times 334000 = 66800 \mathrm{ J}$

Since $Q_{\mathrm{available}} \gt Q_{\mathrm{melt}}$, all the ice melts and the mixture warms above $0^\circ\mathrm{C}$.

Remaining heat after melting:

$Q_{\mathrm{remaining}} = 83720 - 66800 = 16920 \mathrm{ J}$

This heat warms the total mass of water ($0.2 + 0.4 = 0.6$ kg):

$Q_{\mathrm{remaining}} = m_{\mathrm{total}} c_w \Delta T$

$16920 = 0.6 \times 4186 \times T_f$

$T_f = \frac{16920}{2511.6} = 6.7^\circ\mathrm{C}$

The final mixture is all liquid water at $6.7^\circ\mathrm{C}$.

Determining Specific Latent Heat by Electrical Method

Aim: To determine the specific latent heat of fusion of ice (or vaporisation of water) using an electrical heater.

Procedure for latent heat of fusion:

1. Place crushed ice in a funnel with a heating coil immersed in it.
2. Allow the ice to start melting and collect the water that drips through for a few minutes before starting the timer (to ensure the ice is already at $0^\circ\mathrm{C}$).
3. Turn on the heater of known power $P$ for a measured time $t$.
4. Collect the meltwater produced during this time and measure its mass $m$.
5. The specific latent heat of fusion is:

$l_f = \frac{Pt}{m}$

Precautions:

• Use crushed ice to ensure good thermal contact.
• Stir continuously for uniform temperature.
• Account for heat from the surroundings (the heater must also supply the heat that would normally come from the room to melt ice).
• The correct formula accounting for background melting is:

$l_f = \frac{Pt}{m - m_0}$

where $m_0$ is the mass of water collected with the heater off over the same time interval.

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Heat Transfer

Overview

Heat can be transferred by three mechanisms: conduction, convection, and radiation. In most practical situations, more than one mechanism operates simultaneously.

Mechanism	Medium Required	Dominant In	Physical Basis
Conduction	Solid (or stationary fluid)	Metals	Molecular collisions / electron transport
Convection	Fluid (liquid or gas)	Fluids	Bulk motion of fluid
Radiation	None (vacuum)	All, especially high $T$	Electromagnetic waves

Conduction

Mechanism

In conduction, heat is transferred through a material by the vibration and collision of particles, without bulk motion of the material. In metals, free electrons also contribute significantly to heat transfer.

In a non-metal: particles at the hot end vibrate more vigorously and pass energy to neighbours via intermolecular forces. This process is relatively slow.

In a metal: in addition to lattice vibrations, the sea of free electrons can move freely and carry kinetic energy rapidly from the hot end to the cold end. This is why metals are generally much better thermal conductors than non-metals.

Fourier's Law of Heat Conduction

The rate of heat flow through a material is proportional to the temperature gradient and the cross-sectional area:

$\frac{dQ}{dt} = -kA\frac{dT}{dx}$

where:

• $k$ = thermal conductivity ($\mathrm{W m}^{-1}\mathrm{ K}^{-1}$)
• $A$ = cross-sectional area ($\mathrm{m}^2$)
• $\frac{dT}{dx}$ = temperature gradient ($\mathrm{K m}^{-1}$)

The negative sign indicates that heat flows from high to low temperature.

For a uniform slab of thickness $d$ with faces at temperatures $T_1$ (hot) and $T_2$ (cold):

$\frac{Q}{t} = kA\frac{T_1 - T_2}{d}$

Thermal Conductivities

Material	$k$ ($\mathrm{W m}^{-1}\mathrm{ K}^{-1}$)
Copper	$385$
Aluminium	$205$
Steel	$50$
Glass	$0.8$
Water	$0.6$
Brick	$0.6$-$0.8$
Wood	$0.1$-$0.2$
Air	$0.025$
Expanded polystyrene	$0.03$-$0.04$

U-Value (Overall Heat Transfer Coefficient)

The U-value combines the thermal conductivities of all layers in a composite structure (wall, insulation, etc.) into a single figure:

$U = \frac{Q}{A \cdot t \cdot \Delta T}$

Unit: $\mathrm{W m}^{-2}\mathrm{ K}^{-1}$.

For a composite wall with layers of thickness $d_1, d_2, \ldots$ and thermal conductivities $k_1, k_2, \ldots$ (neglecting surface air films):

$\frac{1}{U} = \frac{d_1}{k_1} + \frac{d_2}{k_2} + \cdots = \sum_i \frac{d_i}{k_i}$

The quantity $d_i / k_i$ is called the thermal resistance $R_i$ of layer $i$:

$R_i = \frac{d_i}{k_i}$

$\frac{1}{U} = \sum_i R_i$

A lower U-value indicates better insulation.

Convection

Mechanism

Convection is the transfer of heat by the bulk movement of a fluid. When a fluid is heated, it expands and becomes less dense. The warmer, less dense fluid rises and is replaced by cooler, denser fluid, creating a convection current.

There are two types:

• Natural convection: driven by buoyancy forces due to density differences from temperature differences.
• Forced convection: driven by an external agent (fan, pump, wind).

Examples

• Sea breezes: land heats faster than water during the day; air over land rises, cooler air from the sea flows in to replace it.
• Radiators heat a room primarily by convection (despite the name).
• Atmospheric circulation: differential solar heating drives large-scale convection cells.

Factors Affecting the Rate of Convection

• Temperature difference between the surface and the fluid
• Surface area
• Nature of the fluid (viscosity, thermal expansion coefficient)
• Whether convection is natural or forced

Radiation

Mechanism

Thermal radiation is electromagnetic radiation emitted by all objects with temperature above absolute zero. It does not require a medium and can travel through vacuum.

All objects emit and absorb thermal radiation simultaneously. The net rate of radiative heat transfer depends on the temperature difference between the body and its surroundings.

Stefan-Boltzmann Law

The total power radiated by a black body is:

$P = \sigma A T^4$

where:

• $\sigma = 5.67 \times 10^{-8}$ $\mathrm{W m}^{-2}\mathrm{ K}^{-4}$ (Stefan-Boltzmann constant)
• $A$ = surface area ($\mathrm{m}^2$)
• $T$ = absolute temperature (K)

For a body that is not a perfect black body, we introduce the emissivity $e$ ($0 \le e \le 1$):

$P = e\sigma A T^4$

A perfect black body has $e = 1$. A perfect reflector has $e = 0$. Most dull, dark surfaces have $e \approx 0.8$ to $0.95$; polished, light surfaces have $e \approx 0.1$ to $0.3$.

Net Radiative Power Transfer

For a body at temperature $T_1$ in surroundings at temperature $T_2$ (where $T_1 \gt T_2$):

$P_{\mathrm{net}} = e\sigma A(T_1^4 - T_2^4)$

Black Body Radiation

A black body is an idealised physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. It is also a perfect emitter.

Key features of the black body radiation spectrum:

• The spectrum is continuous and depends only on temperature.
• The total radiated power is proportional to $T^4$ (Stefan-Boltzmann law).
• The wavelength at which the radiation is most intense is inversely proportional to $T$ (Wien's displacement law):

$\lambda_{\mathrm{max}} T = b$

where $b = 2.898 \times 10^{-3}$ $\mathrm{m K}$ (Wien's constant).

• As temperature increases, the peak wavelength shifts to shorter wavelengths (higher frequencies). This explains why objects glow red, then orange, then yellow, then white as they get hotter.

Wien's Displacement Law: Worked Example

The Sun has a surface temperature of approximately $5778$ K. At what wavelength does its radiation peak?

Solution

$\lambda_{\mathrm{max}} = \frac{b}{T} = \frac{2.898 \times 10^{-3}}{5778} = 5.01 \times 10^{-7} \mathrm{ m} = 501 \mathrm{ nm}$

This is in the green-blue region of the visible spectrum, consistent with the Sun appearing yellowish-white (the combined effect of all wavelengths, modified by atmospheric scattering).

Newton's Law of Cooling

When the temperature difference is not too large, the rate of heat loss from a body to its surroundings is approximately proportional to the temperature difference:

$\frac{dQ}{dt} \propto (T_{\mathrm{body}} - T_{\mathrm{surr}})$

Or, for a body of heat capacity $C$:

$C\frac{dT}{dt} = -hA(T - T_s)$

where $h$ is the heat transfer coefficient. This gives exponential decay of the temperature difference:

$T(t) = T_s + (T_0 - T_s)e^{-t/\tau}$

where $\tau = C / (hA)$ is the time constant.

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info radiation is not the dominant mechanism. At large temperature differences, the $T^4$ dependence of radiation becomes significant and the cooling is faster than predicted by Newton's law.

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Gas Laws

Gas Properties

Explore the simulation above to develop intuition for this topic.

Basic Definitions

Quantity	Symbol	Unit	Definition
Pressure	$p$	Pa ($\mathrm{N m}^{-2}$)	Force per unit area
Volume	$V$	$\mathrm{m}^3$	Space occupied by the gas
Temperature	$T$	K	Absolute temperature
Amount of substance	$n$	mol	Number of moles
Molar mass	$M$	$\mathrm{kg mol}^{-1}$	Mass per mole

Standard Temperature and Pressure (STP)

Standard	Temperature	Pressure
STP (IUPAC, 1982)	$273.15$ K ($0^\circ\mathrm{C}$)	$100$ kPa ($1$ bar)
STP (traditional)	$273.15$ K ($0^\circ\mathrm{C}$)	$101.325$ kPa ($1$ atm)

Molar volume at STP (IUPAC): $V_m = 22.7$ L/mol Molar volume at traditional STP: $V_m = 22.4$ L/mol

Boyle's Law

Statement: For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume.

$p \propto \frac{1}{V} \quad (\mathrm{at constant } T, n)$

$p_1 V_1 = p_2 V_2$

Graphical representations:

• $p$ vs $1/V$: straight line through origin (proportionality)
• $p$ vs $V$: rectangular hyperbola (at constant $T$, different curves for different $T$)
• $pV$ vs $p$: horizontal line (at constant $T$)

Charles's Law

Statement: For a fixed mass of gas at constant pressure, the volume is directly proportional to the absolute temperature.

$V \propto T \quad (\mathrm{at constant } p, n)$

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

Critical point: Temperature must be in Kelvin. If Celsius is used, the graph of $V$ vs $T_C$ is a straight line that extrapolates to $V = 0$ at $T_C = -273.15^\circ\mathrm{C}$.

Gay-Lussac's Law (Pressure-Temperature Law)

Statement: For a fixed mass of gas at constant volume, the pressure is directly proportional to the absolute temperature.

$p \propto T \quad (\mathrm{at constant } V, n)$

$\frac{p_1}{T_1} = \frac{p_2}{T_2}$

This is the principle behind the constant-volume gas thermometer and the pressure cooker (the pressure increases as temperature increases at constant volume).

General Gas Law

Combining Boyle's Law and Charles's Law:

$\frac{pV}{T} = \mathrm{constant} \quad (\mathrm{for fixed } n)$

$\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$

Ideal Gas Equation

The ideal gas equation unifies all gas laws:

$pV = nRT$

where $R = 8.314$ $\mathrm{J mol}^{-1}\mathrm{ K}^{-1}$ is the universal molar gas constant.

Alternative forms:

In terms of the number of molecules $N = nN_A$ (where $N_A = 6.022 \times 10^{23}$ $\mathrm{mol}^{-1}$ is Avogadro's number):

$pV = Nk_BT$

where $k_B = R / N_A = 1.381 \times 10^{-23}$ $\mathrm{J K}^{-1}$ is Boltzmann's constant.

In terms of mass and density:

$pV = \frac{m}{M}RT$

$p = \frac{\rho RT}{M}$

where $\rho = m/V$ is the density.

Worked Example 6

A gas occupies $5.0 \times 10^{-3}$ $\mathrm{m}^3$ at $2.0 \times 10^5$ Pa and $300$ K. It is compressed to $2.0 \times 10^{-3}$ $\mathrm{m}^3$ and the pressure increases to $5.0 \times 10^5$ Pa. Find the new temperature.

Solution

$\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$

$T_2 = \frac{p_2 V_2 T_1}{p_1 V_1} = \frac{(5.0 \times 10^5)(2.0 \times 10^{-3})(300)}{(2.0 \times 10^5)(5.0 \times 10^{-3})}$

$T_2 = \frac{3000}{1000} = 300 \mathrm{ K}$

Worked Example 7

Find the number of moles of gas in a $2.0$ L container at $25^\circ\mathrm{C}$ and $1.01 \times 10^5$ Pa.

Solution

$n = \frac{pV}{RT} = \frac{(1.01 \times 10^5)(2.0 \times 10^{-3})}{(8.314)(298.15)}$

$n = \frac{202}{2478.9} = 0.0815 \mathrm{ mol}$

Experimental Verification of Gas Laws

Boyle's Law experiment: A column of air is trapped in a closed tube by a column of oil or mercury. The pressure is varied by changing the height of the oil reservoir, and the volume of the trapped air is measured. A graph of $p$ vs $1/V$ yields a straight line through the origin.

Charles's Law experiment: A capillary tube containing a drop of mercury trapping a column of air is heated in a water bath. The length of the air column (proportional to volume) is measured at various temperatures. A graph of $V$ vs $T$ (in Kelvin) yields a straight line through the origin.

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Ideal Gas Assumptions and Deviations

Assumptions of the Ideal Gas Model

The ideal gas equation is derived under the following assumptions:

1. Point particles: Gas molecules occupy negligible volume compared to the container.
2. No intermolecular forces: Molecules do not exert forces on each other except during collisions.
3. Elastic collisions: Collisions between molecules and between molecules and walls are perfectly elastic (kinetic energy is conserved).
4. Random motion: Molecules move in random directions with a distribution of speeds.
5. Large number of molecules: Statistical treatment is valid.
6. Short duration of collisions: The time of a collision is negligible compared to the time between collisions.

When Do Real Gases Deviate from Ideal Behaviour?

Real gases deviate from ideal behaviour at:

• High pressures: molecules are forced close together, so their volume becomes significant compared to the container volume.
• Low temperatures: molecules move slowly enough that intermolecular attractive forces become significant (this is why gases liquefy at low temperatures).
• High densities: same as high pressure -- molecules are close together.

Under these conditions, the $pV = nRT$ equation gives inaccurate results. The van der Waals equation provides a better model:

$\left(p + \frac{an^2}{V^2}\right)(V - nb) = nRT$

where:

• $a$ accounts for intermolecular attractive forces (reduces effective pressure)
• $b$ accounts for the finite volume of molecules (reduces effective volume)

Compressibility Factor

The compressibility factor $Z$ measures deviation from ideal behaviour:

$Z = \frac{pV}{nRT}$

• $Z = 1$: ideal gas
• $Z \lt 1$: attractive forces dominate (typical at moderate pressures and low temperatures)
• $Z \gt 1$: repulsive forces / molecular volume dominate (typical at very high pressures)

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Kinetic Theory of Gases

Molecular Model

The kinetic theory of gases explains the macroscopic properties of gases (pressure, temperature) in terms of the microscopic behaviour of molecules.

Pressure of an Ideal Gas -- Derivation

Consider $N$ molecules in a cubic container of side $L$. A single molecule of mass $m$ moving with velocity $\mathbf{'\{'}v{'\}'} = (v_x, v_y, v_z)$ collides elastically with a wall perpendicular to the $x$-axis.

Change in momentum per collision:

$\Delta p_x = mv_x - (-mv_x) = 2mv_x$

Time between successive collisions with the same wall:

$\Delta t = \frac{2L}{v_x}$

Force exerted by this molecule on the wall:

$F_x = \frac{\Delta p_x}{\Delta t} = \frac{2mv_x}{2L/v_x} = \frac{mv_x^2}{L}$

Total force from all $N$ molecules:

$F_{\mathrm{total}} = \sum_{i=1}^{N} \frac{mv_{x,i}^2}{L} = \frac{m}{L}\sum_{i=1}^{N} v_{x,i}^2$

Pressure on the wall:

$p = \frac{F_{\mathrm{total}}}{L^2} = \frac{m}{L^3}\sum_{i=1}^{N} v_{x,i}^2 = \frac{mN}{V}\langle v_x^2 \rangle$

where $\langle v_x^2 \rangle = \frac{1}{N}\sum_{i=1}^{N} v_{x,i}^2$ is the mean square velocity in the $x$-direction.

By symmetry, $\langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle$, and:

$\langle v^2 \rangle = \langle v_x^2 \rangle + \langle v_y^2 \rangle + \langle v_z^2 \rangle = 3\langle v_x^2 \rangle$

Therefore:

$p = \frac{mN}{V} \cdot \frac{\langle v^2 \rangle}{3}$

$pV = \frac{1}{3}Nm\langle v^2 \rangle$

This is the fundamental equation of kinetic theory.

Connection to Temperature

Comparing with the ideal gas equation $pV = Nk_BT$:

$\frac{1}{3}Nm\langle v^2 \rangle = Nk_BT$

$\frac{1}{2}m\langle v^2 \rangle = \frac{3}{2}k_BT$

The quantity $\frac{1}{2}m\langle v^2 \rangle$ is the mean translational kinetic energy of a molecule:

$\langle E_k \rangle = \frac{3}{2}k_BT$

This is a central result: temperature is a direct measure of the average translational kinetic energy of molecules.

Root Mean Square Speed

The root mean square (rms) speed is defined as:

$v_{\mathrm{rms}} = \sqrt{\langle v^2 \rangle}$

From the kinetic theory:

$v_{\mathrm{rms}} = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3RT}{M}}$

where $M$ is the molar mass ($M = N_A m$).

Worked Example 8

Find the rms speed of nitrogen molecules ($M = 28.0$ $\mathrm{g/mol} = 0.0280$ $\mathrm{kg/mol}$) at room temperature ($300$ K).

Solution

$v_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.314 \times 300}{0.0280}}$

$v_{\mathrm{rms}} = \sqrt{\frac{7482.6}{0.0280}} = \sqrt{267235.7} = 517 \mathrm{ m/s}$

Maxwell-Boltzmann Speed Distribution

At a given temperature, gas molecules do not all travel at the same speed. The speeds follow the Maxwell-Boltzmann distribution:

$f(v) = 4\pi\left(\frac{m}{2\pi k_BT}\right)^{3/2} v^2 \exp\left(-\frac{mv^2}{2k_BT}\right)$

Key features of the distribution:

• The curve is asymmetric, skewed towards higher speeds (long tail).

• There are three characteristic speeds:

  • Most probable speed $v_p$: speed at the peak of the distribution

    $v_p = \sqrt{\frac{2k_BT}{m}} = \sqrt{\frac{2RT}{M}}$

  • Mean speed $\langle v \rangle$:

    $\langle v \rangle = \sqrt{\frac{8k_BT}{\pi m}} = \sqrt{\frac{8RT}{\pi M}}$

  • Root mean square speed $v_{\mathrm{rms}}$:

    $v_{\mathrm{rms}} = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3RT}{M}}$

• The relationship: $v_p : \langle v \rangle : v_{\mathrm{rms}} = 1 : 1.128 : 1.225$

• As temperature increases, the distribution broadens and shifts to higher speeds.

• As molar mass increases (heavier molecules), the distribution shifts to lower speeds.

Equipartition of Energy

The equipartition theorem states that each degree of freedom that appears quadratically in the energy contributes $\frac{1}{2}k_BT$ to the average energy per molecule (or $\frac{1}{2}RT$ per mole).

For a monatomic ideal gas (3 translational degrees of freedom):

$U = \frac{3}{2}nRT, \quad C_V = \frac{3}{2}R, \quad C_p = \frac{5}{2}R$

For a diatomic ideal gas at moderate temperatures (3 translational + 2 rotational degrees of freedom):

$U = \frac{5}{2}nRT, \quad C_V = \frac{5}{2}R, \quad C_p = \frac{7}{2}R$

The ratio of specific heats:

$\gamma = \frac{C_p}{C_V}$

• Monatomic: $\gamma = 5/3 = 1.667$
• Diatomic: $\gamma = 7/5 = 1.400$

Molar Heat Capacities of Gases

Type	$C_V$ ($\mathrm{J mol}^{-1}\mathrm{ K}^{-1}$)	$C_p$ ($\mathrm{J mol}^{-1}\mathrm{ K}^{-1}$)	$\gamma = C_p/C_V$
Monatomic	$12.5$	$20.8$	$1.67$
Diatomic	$20.8$	$29.1$	$1.40$

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Work Done by an Expanding Gas

Definition

When a gas expands against an external pressure, it does work. For a small expansion $dV$ against pressure $p$:

$dW = p\,dV$

For a finite expansion from volume $V_1$ to $V_2$ at constant pressure:

$W = p(V_2 - V_1) = p\Delta V$

First Law of Thermodynamics

$\Delta U = Q - W$

where:

• $\Delta U$ = change in internal energy (J)
• $Q$ = heat supplied to the system (J)
• $W$ = work done by the system (J)

Sign convention: $Q \gt 0$ when heat is absorbed by the system; $W \gt 0$ when the system does work on the surroundings.

Special Cases

Process	Condition	Work Done	Heat Supplied	Internal Energy Change
Isothermal	$\Delta T = 0$	$W = nRT\ln(V_2/V_1)$	$Q = W$	$\Delta U = 0$
Isochoric (const. $V$)	$\Delta V = 0$	$W = 0$	$Q = nC_V\Delta T$	$\Delta U = Q$
Isobaric (const. $p$)	$\Delta p = 0$	$W = p\Delta V$	$Q = nC_p\Delta T$	$\Delta U = Q - W$
Adiabatic	$Q = 0$	$W = \frac{p_1V_1 - p_2V_2}{\gamma - 1}$	$Q = 0$	$\Delta U = -W$

For an adiabatic process:

$pV^\gamma = \mathrm{constant}$

$TV^{\gamma - 1} = \mathrm{constant}$

Worked Example 9

$2.0$ moles of an ideal monatomic gas expand isothermally at $300$ K from $5.0$ L to $15.0$ L. Find the work done and the heat absorbed.

Solution

$W = nRT\ln\left(\frac{V_2}{V_1}\right) = 2.0 \times 8.314 \times 300 \times \ln\left(\frac{15.0}{5.0}\right)$

$W = 4988.4 \times \ln(3) = 4988.4 \times 1.099 = 5482 \mathrm{ J}$

Since the process is isothermal, $\Delta U = 0$, so $Q = W = 5482$ J.

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Evaporation and Boiling

Evaporation

Evaporation is the process by which molecules escape from the surface of a liquid at temperatures below the boiling point. It occurs because molecules near the surface have a distribution of kinetic energies, and the most energetic ones can overcome the intermolecular forces and escape.

Key features:

• Evaporation occurs only at the surface (unlike boiling, which occurs throughout the liquid).
• Evaporation causes cooling of the remaining liquid (the most energetic molecules leave, lowering the average kinetic energy).
• The rate of evaporation depends on:
  • Temperature (higher temperature = faster evaporation)
  • Surface area (larger area = faster evaporation)
  • Air flow over the surface (removes vapour molecules, maintaining concentration gradient)
  • Vapour pressure of the liquid

Boiling

Boiling occurs when the saturated vapour pressure of the liquid equals the external atmospheric pressure. At this point, bubbles can form within the liquid (not just at the surface) and rise to the top.

• The boiling point of a liquid depends on the external pressure.
• At higher altitudes (lower atmospheric pressure), water boils at a temperature below $100^\circ\mathrm{C}$.
• In a pressure cooker, the high pressure raises the boiling point, allowing food to cook faster.

Distinction Between Evaporation and Boiling

Feature	Evaporation	Boiling
Location	Surface only	Throughout the liquid
Temperature	Any temperature below boiling point	At a specific temperature
Bubbles	No	Yes
Rate	Generally slow	Can be rapid
Condition	Always occurring	Requires SVP $=$ external pressure

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Common Pitfalls

Mistake 1: Confusing Heat and Temperature

Heat and temperature are fundamentally different quantities. A large mass of water at $50^\circ\mathrm{C}$ contains more thermal energy than a small mass of water at $90^\circ\mathrm{C}$. Temperature measures the average kinetic energy per particle; heat is the total energy transferred due to a temperature difference.

Mistake 2: Using Celsius in Gas Law Calculations

All gas law calculations require temperature in Kelvin. Using Celsius will produce incorrect results. For example, doubling the Celsius temperature from $20^\circ\mathrm{C}$ to $40^\circ\mathrm{C}$ is NOT a doubling of the absolute temperature ($293$ K to $313$ K).

Mistake 3: Forgetting to Account for the Calorimeter

In method of mixtures experiments, the calorimeter itself absorbs heat. Ignoring the calorimeter's heat capacity $C_{\mathrm{cal}}$ leads to an overestimate of the specific heat capacity of the sample. The correct energy balance is:

$m_s c_s (T_s - T_f) = (m_w c_w + C_{\mathrm{cal}})(T_f - T_w)$

Mistake 4: Assuming All Ice Melts (or All Water Freezes)

When mixing ice and water, always check whether the available heat is sufficient to melt all the ice before assuming the final state is all liquid. If not, the final temperature is $0^\circ\mathrm{C}$ and the system is a mixture of ice and water.

Mistake 5: Wrong Sign Convention in the First Law

The first law of thermodynamics is $\Delta U = Q - W$. Here $W$ is work done by the system. If the problem gives work done on the system, you must negate it. Be consistent with the sign convention throughout.

Mistake 6: Confusing the Three Characteristic Speeds

The most probable speed, mean speed, and rms speed are different. For Maxwell-Boltzmann distributions:

$v_p = \sqrt{\frac{2RT}{M}}, \quad \langle v \rangle = \sqrt{\frac{8RT}{\pi M}}, \quad v_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}}$

The relationship is $v_p \lt \langle v \rangle \lt v_{\mathrm{rms}}$. In DSE problems, pay attention to which speed the question asks for.

Mistake 7: Applying Boyle's Law When Temperature Changes

Boyle's Law ($p_1 V_1 = p_2 V_2$) is only valid when temperature is constant. If the problem involves a temperature change, you must use the general gas law:

$\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$

Mistake 8: Units in Thermal Conductivity Problems

When using Fourier's law, ensure all quantities are in SI units: $k$ in $\mathrm{W m}^{-1}\mathrm{ K}^{-1}$, $A$ in $\mathrm{m}^2$, $d$ in $\mathrm{m}$, $\Delta T$ in K. A common error is using centimetres for thickness without converting to metres.

Mistake 9: Mixing Up Specific Heat Capacity and Heat Capacity

Specific heat capacity $c$ is per unit mass ($\mathrm{J kg}^{-1}\mathrm{ K}^{-1}$). Heat capacity $C$ is for the entire object ($\mathrm{J K}^{-1}$). They are related by $C = mc$. Using one in place of the other is a frequent error.

Mistake 10: Incorrect U-Value Calculation

For a composite wall, thermal resistances add in series (not conductances). The correct formula is:

$\frac{1}{U} = \frac{d_1}{k_1} + \frac{d_2}{k_2} + \cdots$

Do not add the U-values of individual layers directly.

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Practice Problems

Question 1: Temperature Conversion

The surface temperature of the Sun is approximately $5778$ K. Express this in Celsius and Fahrenheit.

Solution

$T_C = 5778 - 273.15 = 5504.85^\circ\mathrm{C}$

$T_F = \frac{9}{5}(5504.85) + 32 = 9908.73 + 32 = 9940.73^\circ\mathrm{F}$

Question 2: Specific Heat Capacity Mixing

A $0.3$ kg aluminium block at $200^\circ\mathrm{C}$ is dropped into $1.0$ kg of oil at $25^\circ\mathrm{C}$ contained in a copper calorimeter of mass $0.2$ kg at $25^\circ\mathrm{C}$. The specific heat capacity of oil is $2100$ $\mathrm{J kg}^{-1}\mathrm{ K}^{-1}$. Find the final temperature. (Specific heat capacity of aluminium $= 900$ $\mathrm{J kg}^{-1}\mathrm{ K}^{-1}$, copper $= 385$ $\mathrm{J kg}^{-1}\mathrm{ K}^{-1}$.)

Solution

Heat lost by aluminium $=$ heat gained by oil $+$ heat gained by calorimeter:

$m_a c_a (T_a - T_f) = m_o c_o (T_f - T_o) + m_c c_c (T_f - T_o)$

$0.3 \times 900 \times (200 - T_f) = 1.0 \times 2100 \times (T_f - 25) + 0.2 \times 385 \times (T_f - 25)$

$270(200 - T_f) = 2100(T_f - 25) + 77(T_f - 25)$

$54000 - 270T_f = 2177T_f - 54425$

$54000 + 54425 = 2177T_f + 270T_f$

$108425 = 2447T_f$

$T_f = 44.3^\circ\mathrm{C}$

Question 3: Latent Heat with Phase Change

$100$ g of steam at $100^\circ\mathrm{C}$ is passed into $500$ g of water at $20^\circ\mathrm{C}$. Find the final temperature and state.

Solution

Step 1: Check if all steam condenses.

Heat released if all steam condenses and cools to $0^\circ\mathrm{C}$:

$Q_{\mathrm{max}} = m_s l_v + m_s c_w (100 - 0) = 0.1 \times 2260000 + 0.1 \times 4186 \times 100$

$Q_{\mathrm{max}} = 226000 + 41860 = 267860 \mathrm{ J}$

Heat required to warm water from $20^\circ\mathrm{C}$ to $100^\circ\mathrm{C}$:

$Q_{\mathrm{warm}} = m_w c_w (100 - 20) = 0.5 \times 4186 \times 80 = 167440 \mathrm{ J}$

Since $Q_{\mathrm{max}} \gt Q_{\mathrm{warm}}$, all the steam condenses and the final temperature is above $20^\circ\mathrm{C}$ but we need to check if it reaches $100^\circ\mathrm{C}$.

Actually, $Q_{\mathrm{warm}} = 167440$ J is less than the latent heat of condensation alone ($226000$ J), so only part of the steam condenses. The final temperature is $100^\circ\mathrm{C}$.

Let $m$ be the mass of steam that condenses:

$m \times l_v = 167440$

$m = \frac{167440}{2260000} = 0.0741 \mathrm{ kg} = 74.1 \mathrm{ g}$

The final mixture is $174.1$ g of water and $25.9$ g of steam at $100^\circ\mathrm{C}$.

Question 4: Thermal Conductivity

A glass window of area $1.5$ $\mathrm{m}^2$ and thickness $5.0$ mm has an indoor surface temperature of $20^\circ\mathrm{C}$ and an outdoor surface temperature of $5^\circ\mathrm{C}$. Find the rate of heat loss through the window. (Thermal conductivity of glass $= 0.8$ $\mathrm{W m}^{-1}\mathrm{ K}^{-1}$.)

Solution

$\frac{Q}{t} = kA\frac{T_1 - T_2}{d} = 0.8 \times 1.5 \times \frac{20 - 5}{0.005}$

$\frac{Q}{t} = 1.2 \times \frac{15}{0.005} = 1.2 \times 3000 = 3600 \mathrm{ W}$

Question 5: U-Value of Composite Wall

A wall consists of a $10$ cm brick layer ($k = 0.6$ $\mathrm{W m}^{-1}\mathrm{ K}^{-1}$) and a $5$ cm layer of insulation ($k = 0.04$ $\mathrm{W m}^{-1}\mathrm{ K}^{-1}$). Find the U-value.

Solution

$R_{\mathrm{brick}} = \frac{d_1}{k_1} = \frac{0.10}{0.6} = 0.1667 \mathrm{ m}^2\mathrm{ K/W}$

$R_{\mathrm{insulation}} = \frac{d_2}{k_2} = \frac{0.05}{0.04} = 1.25 \mathrm{ m}^2\mathrm{ K/W}$

$\frac{1}{U} = 0.1667 + 1.25 = 1.4167 \mathrm{ m}^2\mathrm{ K/W}$

$U = \frac{1}{1.4167} = 0.706 \mathrm{ W m}^{-2}\mathrm{ K}^{-1}$

Question 6: Stefan-Boltzmann Law

A spherical black body of radius $5.0$ cm is maintained at $500$ K. Find the power radiated.

Solution

$P = \sigma A T^4 = \sigma \times 4\pi r^2 \times T^4$

$P = 5.67 \times 10^{-8} \times 4\pi(0.05)^2 \times (500)^4$

$P = 5.67 \times 10^{-8} \times 0.03142 \times 6.25 \times 10^{10}$

$P = 5.67 \times 10^{-8} \times 1.963 \times 10^9 = 111.3 \mathrm{ W}$

Question 7: Gas Law Combined

A gas cylinder contains $10$ L of oxygen at $20^\circ\mathrm{C}$ and $1.5 \times 10^6$ Pa. If the temperature rises to $80^\circ\mathrm{C}$ and the pressure valve releases gas to maintain $1.5 \times 10^6$ Pa, what volume of gas (measured at $20^\circ\mathrm{C}$ and $1.01 \times 10^5$ Pa) escapes?

Solution

The gas that remains in the cylinder at $80^\circ\mathrm{C}$ and $1.5 \times 10^6$ Pa occupies $10$ L.

Moles remaining:

$n_r = \frac{pV}{RT} = \frac{1.5 \times 10^6 \times 0.010}{8.314 \times 353.15} = \frac{15000}{2936.7} = 5.108 \mathrm{ mol}$

Initial moles:

$n_i = \frac{1.5 \times 10^6 \times 0.010}{8.314 \times 293.15} = \frac{15000}{2437.5} = 6.154 \mathrm{ mol}$

Moles escaped:

$\Delta n = 6.154 - 5.108 = 1.046 \mathrm{ mol}$

Volume at STP conditions ($20^\circ\mathrm{C}$, $1.01 \times 10^5$ Pa):

$V = \frac{nRT}{p} = \frac{1.046 \times 8.314 \times 293.15}{1.01 \times 10^5} = \frac{2548.5}{101000} = 0.0252 \mathrm{ m}^3 = 25.2 \mathrm{ L}$

Question 8: RMS Speed Comparison

Compare the rms speeds of hydrogen ($M = 2.0$ $\mathrm{g/mol}$) and oxygen ($M = 32.0$ $\mathrm{g/mol}$) at the same temperature.

Solution

$\frac{v_{\mathrm{rms,H}_2}}{v_{\mathrm{rms,O}_2}} = \sqrt{\frac{M_{\mathrm{O}_2}}{M_{\mathrm{H}_2}}} = \sqrt{\frac{32.0}{2.0}} = \sqrt{16} = 4$

Hydrogen molecules move four times faster than oxygen molecules at the same temperature. This is consistent with the observation that lighter gases diffuse more rapidly (Graham's law).

Question 9: First Law of Thermodynamics

$1.0$ mol of an ideal monatomic gas at $300$ K expands isobarically from $10$ L to $20$ L. Find the work done, the change in internal energy, and the heat supplied.

Solution

Work done:

$W = p\Delta V$

First, find $p$:

$p = \frac{nRT}{V_1} = \frac{1.0 \times 8.314 \times 300}{0.010} = 249420 \mathrm{ Pa}$

$W = 249420 \times (0.020 - 0.010) = 2494.2 \mathrm{ J}$

Change in internal energy (for a monatomic ideal gas, $C_V = \frac{3}{2}R$):

Find the final temperature:

$\frac{V_1}{T_1} = \frac{V_2}{T_2} \implies T_2 = \frac{V_2 T_1}{V_1} = \frac{20 \times 300}{10} = 600 \mathrm{ K}$

$\Delta U = nC_V\Delta T = 1.0 \times \frac{3}{2} \times 8.314 \times (600 - 300) = 12.471 \times 300 = 3741.3 \mathrm{ J}$

Heat supplied:

$Q = \Delta U + W = 3741.3 + 2494.2 = 6235.5 \mathrm{ J}$

Alternatively, using $C_p = \frac{5}{2}R$:

$Q = nC_p\Delta T = 1.0 \times \frac{5}{2} \times 8.314 \times 300 = 6235.5 \mathrm{ J}$

Question 10: Adiabatic Expansion

$2.0$ mol of a diatomic ideal gas ($\gamma = 1.4$) expands adiabatically from $V_1 = 5.0$ L at $p_1 = 4.0 \times 10^5$ Pa to $V_2 = 10.0$ L. Find the final pressure and temperature.

Solution

Final pressure:

$p_1 V_1^\gamma = p_2 V_2^\gamma$

$p_2 = p_1\left(\frac{V_1}{V_2}\right)^\gamma = 4.0 \times 10^5 \times \left(\frac{5.0}{10.0}\right)^{1.4}$

$p_2 = 4.0 \times 10^5 \times (0.5)^{1.4} = 4.0 \times 10^5 \times 0.3789 = 1.516 \times 10^5 \mathrm{ Pa}$

Initial temperature:

$T_1 = \frac{p_1 V_1}{nR} = \frac{4.0 \times 10^5 \times 0.0050}{2.0 \times 8.314} = \frac{2000}{16.628} = 120.3 \mathrm{ K}$

Final temperature:

$T_2 = \frac{p_2 V_2}{nR} = \frac{1.516 \times 10^5 \times 0.010}{2.0 \times 8.314} = \frac{1516}{16.628} = 91.1 \mathrm{ K}$

The temperature drops because the gas does work on its surroundings without any heat input.

Question 11: Newton's Law of Cooling

A body cools from $80^\circ\mathrm{C}$ to $60^\circ\mathrm{C}$ in $5$ minutes in a room at $20^\circ\mathrm{C}$. How long will it take to cool from $60^\circ\mathrm{C}$ to $40^\circ\mathrm{C}$?

Solution

Using Newton's law of cooling:

$T(t) = T_s + (T_0 - T_s)e^{-t/\tau}$

For the first interval ($80^\circ\mathrm{C}$ to $60^\circ\mathrm{C}$ in $300$ s):

$60 = 20 + (80 - 20)e^{-300/\tau}$

$40 = 60e^{-300/\tau}$

$e^{-300/\tau} = \frac{2}{3}$

$\frac{-300}{\tau} = \ln\left(\frac{2}{3}\right)$

$\tau = \frac{-300}{\ln(2/3)} = \frac{-300}{-0.4055} = 739.8 \mathrm{ s}$

For the second interval ($60^\circ\mathrm{C}$ to $40^\circ\mathrm{C}$):

$40 = 20 + (60 - 20)e^{-t/\tau}$

$20 = 40e^{-t/\tau}$

$e^{-t/\tau} = \frac{1}{2}$

$\frac{-t}{739.8} = \ln\left(\frac{1}{2}\right) = -0.693$

$t = 739.8 \times 0.693 = 512.7 \mathrm{ s} \approx 8.5 \mathrm{ minutes}$

Question 12: Wien's Displacement Law

A star has a surface temperature of $3500$ K. (a) At what wavelength does it radiate most intensely? (b) In what region of the electromagnetic spectrum is this?

Solution

$\lambda_{\mathrm{max}} = \frac{b}{T} = \frac{2.898 \times 10^{-3}}{3500} = 8.28 \times 10^{-7} \mathrm{ m} = 828 \mathrm{ nm}$

This is in the near-infrared region (visible light extends from approximately $380$ nm to $750$ nm). Such stars appear reddish to the eye because the tail of the distribution extends into the red part of the visible spectrum.

Question 13: Mean Free Path (Extension)

Oxygen molecules at STP have a molecular diameter of $3.6 \times 10^{-10}$ m. Estimate the mean free path.

Solution

The mean free path is:

$\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$

where $n$ is the number density. At STP, $n = \frac{p}{k_BT}$:

$n = \frac{1.01 \times 10^5}{1.381 \times 10^{-23} \times 273.15} = 2.68 \times 10^{25} \mathrm{ m}^{-3}$

$\lambda = \frac{1}{\sqrt{2}\pi(3.6 \times 10^{-10})^2 \times 2.68 \times 10^{25}}$

$\lambda = \frac{1}{1.414 \times 3.14159 \times 1.296 \times 10^{-19} \times 2.68 \times 10^{25}}$

$\lambda = \frac{1}{1.537 \times 10^{7}} = 6.51 \times 10^{-8} \mathrm{ m} = 65.1 \mathrm{ nm}$

Question 14: Density of a Gas

Find the density of nitrogen gas ($M = 28.0$ $\mathrm{g/mol}$) at $25^\circ\mathrm{C}$ and $1.01 \times 10^5$ Pa.

Solution

$\rho = \frac{pM}{RT} = \frac{(1.01 \times 10^5)(0.0280)}{(8.314)(298.15)}$

$\rho = \frac{2828}{2478.9} = 1.141 \mathrm{ kg/m}^3$

Question 15: Complete Energy Balance with Phase Change

A $500$ W heater is used to heat $0.5$ kg of ice at $-10^\circ\mathrm{C}$ in an insulated container. How long does it take to convert all the ice to steam at $100^\circ\mathrm{C}$? Assume $100\%$ efficiency.

Solution

Stage 1: Heat ice from $-10^\circ\mathrm{C}$ to $0^\circ\mathrm{C}$:

$Q_1 = 0.5 \times 2100 \times 10 = 10500 \mathrm{ J}$

Stage 2: Melt ice at $0^\circ\mathrm{C}$:

$Q_2 = 0.5 \times 334000 = 167000 \mathrm{ J}$

Stage 3: Heat water from $0^\circ\mathrm{C}$ to $100^\circ\mathrm{C}$:

$Q_3 = 0.5 \times 4186 \times 100 = 209300 \mathrm{ J}$

Stage 4: Vaporise water at $100^\circ\mathrm{C}$:

$Q_4 = 0.5 \times 2260000 = 1130000 \mathrm{ J}$

Total energy:

$Q_{\mathrm{total}} = 10500 + 167000 + 209300 + 1130000 = 1516800 \mathrm{ J}$

Time:

$t = \frac{Q_{\mathrm{total}}}{P} = \frac{1516800}{500} = 3033.6 \mathrm{ s} = 50.6 \mathrm{ minutes}$

For the A-Level treatment of this topic, see Thermal Properties.

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tip

tip Ready to test your understanding of Heat and Gases? The diagnostic test contains the hardest questions within the DSE specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Heat and Gases with other physics topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.

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Derivations

Derivation: Ideal Gas Law from Empirical Gas Laws

Boyle's law (constant temperature): $PV = \text{constant}$, or $P \propto 1/V$

Charles's law (constant pressure): $V/T = \text{constant}$, or $V \propto T$

Pressure law (constant volume): $P/T = \text{constant}$, or $P \propto T$

Combining all three: $PV/T = \text{constant}$ for a fixed mass of gas.

For $n$ moles of ideal gas:

$PV = nRT$

where $R = 8.31 \mathrm{ J/(mol\cdot K)}$ is the molar gas constant.

For one mole: $PV = RT$. At STP ($T = 273 \mathrm{ K}$, $P = 1.01 \times 10^5 \mathrm{ Pa}$): $V = RT/P = 8.31 \times 273 / (1.01 \times 10^5) = 0.0224 \mathrm{ m}^3 = 22.4 \mathrm{ litres}$.

Derivation: Kinetic Theory Pressure Formula

Consider $N$ molecules in a cubic container of side $L$. A molecule of mass $m$ moving with velocity component $u_x$ in the x-direction bounces off a wall. The change in momentum per collision is $\Delta p = 2mu_x$. The time between collisions with the same wall is $2L/u_x$.

Force on wall from one molecule: $F = \frac{\Delta p}{\Delta t} = \frac{2mu_x}{2L/u_x} = \frac{mu_x^2}{L}$

Total force on wall from all $N$ molecules: $F = \frac{m}{L}\sum_{i=1}^{N} u_{xi}^2$

Pressure: $P = \frac{F}{L^2} = \frac{m}{L^3}\sum u_{xi}^2 = \frac{mN\overline{u_x^2}}{V}$

Since $\overline{u^2} = \overline{u_x^2} + \overline{u_y^2} + \overline{u_z^2} = 3\overline{u_x^2}$ (random motion):

$PV = \frac{1}{3}Nm\overline{u^2} = \frac{2}{3}N \times \frac{1}{2}m\overline{u^2} = \frac{2}{3}N \times \overline{E_k}$

This is the kinetic theory equation: $PV = \frac{2}{3}N\overline{E_k}$.

Derivation: Root-Mean-Square Speed

From the kinetic theory equation and ideal gas law:

$PV = nRT = \frac{2}{3}N\overline{E_k} = \frac{2}{3}N \times \frac{1}{2}m\overline{u^2} = \frac{1}{3}Nm\overline{u^2}$

$\overline{u^2} = \frac{3RT}{M}$

where $M = Nm/N_A = mN_A$ is the molar mass.

The root-mean-square speed: $u_{\mathrm{rms}} = \sqrt{\overline{u^2}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3k_BT}{m}}$

where $k_B = R/N_A = 1.38 \times 10^{-23} \mathrm{ J/K}$ is Boltzmann's constant.

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Experimental Methods

Determining the Specific Heat Capacity of a Metal

Apparatus: A metal block (e.g., aluminium) with two holes (for thermometer and heater), an immersion heater, a thermometer, an ammeter, a voltmeter, a stopwatch, and insulation.

Procedure:

1. Measure the mass $m$ of the metal block.
2. Insert the heater and thermometer, and insulate the block.
3. Record the initial temperature $T_i$.
4. Switch on the heater, record the voltage $V$ and current $I$.
5. After time $t$, record the final temperature $T_f$.
6. Calculate: $c = \frac{VIt}{m(T_f - T_i)}$

Sources of error:

• Heat loss to the surroundings despite insulation.
• Non-uniform temperature within the block.
• The heater and thermometer have their own heat capacities.

Improvements: Plot temperature versus time, extrapolate the cooling portion back to estimate the temperature that would have been reached without heat loss. Repeat and average.

Verifying Boyle's Law

Apparatus: A column of air trapped in a sealed glass tube by a column of oil (or mercury), connected to a pressure gauge or manometer.

Procedure:

1. Record the pressure $P$ and the length $L$ (proportional to volume) of the trapped air.
2. Vary the pressure by adjusting the oil reservoir.
3. Record pairs of $P$ and $L$.
4. Plot $P$ (y-axis) versus $1/L$ (x-axis). A straight line through the origin confirms Boyle's law ($P \propto 1/V$).
5. Alternatively, plot $PV$ (y-axis) versus $P$ (x-axis). A horizontal line confirms $PV = \text{const}$.

Precautions:

• Allow time for the gas to reach thermal equilibrium after each pressure change.
• Keep the temperature constant throughout.

Investigating the Pressure Law

Apparatus: A flask of air connected to a pressure gauge, immersed in a water bath with a thermometer.

Procedure:

1. Heat the water bath gradually and record pairs of pressure $P$ and temperature $T$ (in kelvin).
2. Plot $P$ (y-axis) versus $T$ (x-axis).
3. A straight line through the origin confirms $P \propto T$ (pressure law).

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Data Analysis and Uncertainty

Uncertainty in Specific Heat Capacity

For $c = \frac{VIt}{m\Delta T}$:

$\frac{\Delta c}{c} = \sqrt{\left(\frac{\Delta V}{V}\right)^2 + \left(\frac{\Delta I}{I}\right)^2 + \left(\frac{\Delta t}{t}\right)^2 + \left(\frac{\Delta m}{m}\right)^2 + \left(\frac{\Delta(\Delta T)}{\Delta T}\right)^2}$

The temperature difference uncertainty is: $\Delta(\Delta T) = \sqrt{(\Delta T_f)^2 + (\Delta T_i)^2}$

Example: $V = (12.0 \pm 0.1) \mathrm{ V}$, $I = (2.00 \pm 0.02) \mathrm{ A}$, $t = (300 \pm 1) \mathrm{ s}$, $m = (0.500 \pm 0.001) \mathrm{ kg}$, $\Delta T = (25.0 \pm 0.5)^\circ\mathrm{C}$:

c = \frac{12.0 \times 2.00 \times 300}{0.500 \times 25.0} = \frac{7200}{12.5} = 576 \mathrm{ J/(kg\cdot}^\circ C)}

$\frac{\Delta c}{c} = \sqrt{(0.0083)^2 + (0.01)^2 + (0.0033)^2 + (0.002)^2 + (0.02)^2} = \sqrt{0.000069 + 0.0001 + 0.000011 + 0.000004 + 0.0004} = \sqrt{0.000584} = 0.0242 = 2.4\%$

\Delta c = 0.024 \times 576 = 14 \mathrm{ J/(kg\cdot}^\circ C)}

c = (576 \pm 14) \mathrm{ J/(kg\cdot}^\circ C)}

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Additional Worked Examples

Worked Example 11

A gas in a cylinder with a movable piston is compressed isothermally at $300 \mathrm{ K}$ from volume $5.0 \times 10^{-3} \mathrm{ m}^3$ at pressure $1.0 \times 10^5 \mathrm{ Pa}$ to volume $2.0 \times 10^{-3} \mathrm{ m}^3$. Calculate the final pressure and the work done on the gas.

Solution

By Boyle's law: $P_1 V_1 = P_2 V_2$

$P_2 = \frac{P_1 V_1}{V_2} = \frac{1.0 \times 10^5 \times 5.0 \times 10^{-3}}{2.0 \times 10^{-3}} = 2.5 \times 10^5 \mathrm{ Pa}$

Work done on the gas during isothermal compression:

$W = nRT\ln\left(\frac{V_2}{V_1}\right)$

Since $nRT = P_1 V_1 = 1.0 \times 10^5 \times 5.0 \times 10^{-3} = 500 \mathrm{ J}$:

$W = 500 \times \ln\left(\frac{2.0}{5.0}\right) = 500 \times \ln(0.4) = 500 \times (-0.916) = -458 \mathrm{ J}$

The negative sign means work is done on the gas (volume decreases). The magnitude is $458 \mathrm{ J}$.

Worked Example 12

A room measures $5.0 \mathrm{ m} \times 4.0 \mathrm{ m} \times 3.0 \mathrm{ m}$ and contains air at $20^\circ\mathrm{C}$ and $1.01 \times 10^5 \mathrm{ Pa}$.

(a) Calculate the number of moles of air in the room.

(b) Calculate the number of air molecules.

(c) If the temperature rises to $25^\circ\mathrm{C}$ at constant pressure, calculate the mass of air that leaves the room.

(Molar mass of air $= 29 \mathrm{ g/mol}$)

Solution

(a) Volume: $V = 5.0 \times 4.0 \times 3.0 = 60.0 \mathrm{ m}^3$

$n = \frac{PV}{RT} = \frac{1.01 \times 10^5 \times 60.0}{8.31 \times 293} = \frac{6.06 \times 10^6}{2434.8} = 2489 \mathrm{ mol}$

(b) $N = n \times N_A = 2489 \times 6.02 \times 10^{23} = 1.50 \times 10^{27}$ molecules

(c) At constant pressure, $V \propto T$. The volume at $25^\circ\mathrm{C}$ ($298 \mathrm{ K}$):

$V_2 = V_1 \times \frac{T_2}{T_1} = 60.0 \times \frac{298}{293} = 61.0 \mathrm{ m}^3$

The room volume is fixed at $60.0 \mathrm{ m}^3$, so the excess air ($1.0 \mathrm{ m}^3$ at $298 \mathrm{ K}$) leaves.

Moles leaving: $n_{\mathrm{out}} = n_1 \times \left(1 - \frac{T_1}{T_2}\right) = 2489 \times \left(1 - \frac{293}{298}\right) = 2489 \times 0.0168 = 41.8 \mathrm{ mol}$

Mass leaving: $m = 41.8 \times 0.029 = 1.21 \mathrm{ kg}$

Worked Example 13

Calculate the root-mean-square speed of nitrogen molecules ($M = 28 \mathrm{ g/mol}$) at: (a) $0^\circ\mathrm{C}$, (b) $100^\circ\mathrm{C}$.

Solution

$u_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}}$

(a) At $0^\circ\mathrm{C}$ ($273 \mathrm{ K}$):

$u_{\mathrm{rms}} = \sqrt{\frac{3 \times 8.31 \times 273}{0.028}} = \sqrt{\frac{6806.3}{0.028}} = \sqrt{243080} = 493 \mathrm{ m/s}$

(b) At $100^\circ\mathrm{C}$ ($373 \mathrm{ K}$):

$u_{\mathrm{rms}} = \sqrt{\frac{3 \times 8.31 \times 373}{0.028}} = \sqrt{\frac{9298.6}{0.028}} = \sqrt{332093} = 576 \mathrm{ m/s}$

Note: $u_{\mathrm{rms}}$ increases with temperature but depends on the square root of $T$, so doubling the absolute temperature only increases $u_{\mathrm{rms}}$ by a factor of $\sqrt{2} \approx 1.41$.

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Exam-Style Questions

Question 1 (DSE Structured)

(a) State the assumptions of the kinetic theory of gases.

(b) A gas cylinder contains $2.0 \mathrm{ mol}$ of an ideal gas at temperature $300 \mathrm{ K}$ and pressure $2.0 \times 10^6 \mathrm{ Pa}$.

(i) Calculate the volume of the gas.

(ii) Calculate the total kinetic energy of the gas molecules.

(iii) Calculate the root-mean-square speed of the molecules if the molar mass is $4.0 \mathrm{ g/mol}$.

(c) Explain why real gases deviate from ideal gas behaviour at high pressures and low temperatures.

Solution

(a) Assumptions of the kinetic theory:

1. The gas consists of a large number of small molecules in random motion.
2. Collisions between molecules and with the walls are perfectly elastic.
3. The volume of the molecules is negligible compared to the volume of the container.
4. Intermolecular forces (other than during collisions) are negligible.
5. The time spent in collisions is negligible compared to the time between collisions.

(b) (i) $V = \frac{nRT}{P} = \frac{2.0 \times 8.31 \times 300}{2.0 \times 10^6} = \frac{4986}{2.0 \times 10^6} = 2.49 \times 10^{-3} \mathrm{ m}^3$

(ii) From $PV = \frac{2}{3}N\overline{E_k}$: $\overline{E_k} = \frac{3PV}{2N} = \frac{3PV}{2nN_A}$

Alternatively, $\overline{E_k} = \frac{3}{2}k_BT = \frac{3}{2} \times \frac{R}{N_A} \times T = \frac{3RT}{2N_A}$

Total KE $= N \times \overline{E_k} = nN_A \times \frac{3RT}{2N_A} = \frac{3}{2}nRT = \frac{3}{2} \times 2.0 \times 8.31 \times 300 = 7479 \mathrm{ J}$

(iii) $u_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.31 \times 300}{0.004}} = \sqrt{\frac{7479}{0.004}} = \sqrt{1.87 \times 10^6} = 1367 \mathrm{ m/s}$

(c) At high pressures, molecules are forced close together, so the volume of the molecules themselves becomes significant compared to the container volume (assumption 3 fails). At low temperatures, molecules move more slowly, so intermolecular forces become significant (assumption 4 fails), and the gas may even liquefy.

Question 2 (DSE Structured)

A student carries out an experiment to determine the specific heat capacity of water. She heats $0.200 \mathrm{ kg}$ of water using an immersion heater of power $50 \mathrm{ W}$. The following data are recorded:

Time (s)	Temperature ($^\circ\mathrm{C}$)
0	20.0
60	23.5
120	27.2
180	30.8
240	34.7
300	38.3

(a) Plot a graph of temperature versus time and determine the rate of temperature rise.

(b) Calculate the specific heat capacity of water from the gradient.

(c) The accepted value is 4186 \mathrm{ J/(kg\cdot}^\circ C)}. Calculate the percentage error and suggest two reasons for any discrepancy.

Solution

(a) The graph of temperature versus time is approximately linear. The gradient (rate of temperature rise) from a line of best fit:

$\mathrm{Gradient} = \frac{\Delta T}{\Delta t} \approx \frac{38.3 - 20.0}{300 - 0} = \frac{18.3}{300} = 0.0610^\circ\mathrm{C/s}$

(b) P = mc \times \frac{\Delta T}{\Delta t} \implies c = \frac{P}{m \times \mathrm{gradient}} = \frac{50}{0.200 \times 0.0610} = \frac{50}{0.0122} = 4100 \mathrm{ J/(kg\cdot}^\circ C)}

(c) Percentage error $= \frac{|4100 - 4186|}{4186} \times 100\% = \frac{86}{4186} \times 100\% = 2.1\%$

Two reasons for discrepancy:

1. Heat loss to the surroundings: Some electrical energy heats the container and the air, not just the water.
2. Incomplete stirring: The temperature may not be uniform, so the recorded temperature may not represent the average water temperature.

Question 3 (DSE Structured)

(a) Distinguish between evaporation and boiling.

(b) $0.050 \mathrm{ kg}$ of ice at $-10^\circ\mathrm{C}$ is added to $0.300 \mathrm{ kg}$ of water at $40^\circ\mathrm{C}$ in an insulated container. Calculate the final temperature of the mixture.

(Specific heat capacity of ice = 2100 \mathrm{ J/(kg\cdot}^\circ C)}, specific latent heat of fusion of ice $= 334000 \mathrm{ J/kg}$, specific heat capacity of water = 4200 \mathrm{ J/(kg\cdot}^\circ C)})

Solution

(a) Evaporation occurs at any temperature, only at the surface, and is a slow process. It causes cooling because the most energetic molecules escape. Boiling occurs at a specific temperature (the boiling point), throughout the liquid, and is a rapid process involving bubble formation. Both involve a change of state from liquid to gas.

(b) First, check if all the ice melts. Energy to warm ice to $0^\circ\mathrm{C}$:

$Q_1 = 0.050 \times 2100 \times 10 = 1050 \mathrm{ J}$

Energy to melt ice at $0^\circ\mathrm{C}$:

$Q_2 = 0.050 \times 334000 = 16700 \mathrm{ J}$

Total energy needed to melt all ice: $Q_1 + Q_2 = 1050 + 16700 = 17750 \mathrm{ J}$

Maximum energy available from water cooling to $0^\circ\mathrm{C}$:

$Q_{\max} = 0.300 \times 4200 \times 40 = 50400 \mathrm{ J}$

Since $50400 > 17750$, all the ice melts and the final temperature is above $0^\circ\mathrm{C}$.

Energy available after melting ice: $50400 - 17750 = 32650 \mathrm{ J}$

This energy warms the melted ice and cools the original water:

$(0.050 \times 4200 + 0.300 \times 4200) \times \Delta T = 32650$

$4200(0.050 + 0.300) \times \Delta T = 32650$

$4200 \times 0.350 \times \Delta T = 32650$

$1470\Delta T = 32650$

$\Delta T = 22.2^\circ\mathrm{C}$

Final temperature $= 0 + 22.2 = 22.2^\circ\mathrm{C}$.

(Alternatively: $0.300 \times 4200 \times (40 - T_f) = 0.050 \times 2100 \times 10 + 0.050 \times 334000 + 0.050 \times 4200 \times T_f$)

$50400 - 1260T_f = 1050 + 16700 + 210T_f$

$50400 - 17750 = 1470T_f$

$$T_f = 22.2^\circ\mathrm\\{C\\}$