Heat and Gases — Diagnostic Tests

Unit Tests

UT-1: Multi-Stage Heating with Phase Change

Question:

A $200$ g block of ice at $-15°$C is heated by a $500$ W heater until it becomes steam at $110°$C. The specific heat capacity of ice is $2100 \text{ J kg}^{-1}\text{K}^{-1}$, the specific heat capacity of water is $4200 \text{ J kg}^{-1}\text{K}^{-1}$, the specific heat capacity of steam is $2010 \text{ J kg}^{-1}\text{K}^{-1}$, the specific latent heat of fusion of ice is $3.34 \times 10^5 \text{ J kg}^{-1}$, and the specific latent heat of vaporisation of water is $2.26 \times 10^6 \text{ J kg}^{-1}$. Find (a) the total energy required, and (b) the total time taken.

Solution:

(a) Total energy (five stages):

Stage 1: Ice from $-15°$C to $0°$C:

$Q_1 = mc_{\text{ice}}\Delta T = 0.2 \times 2100 \times 15 = 6300 \text{ J}$

Stage 2: Melting ice at $0°$C:

$Q_2 = mL_f = 0.2 \times 3.34 \times 10^5 = 66800 \text{ J}$

Stage 3: Water from $0°$C to $100°$C:

$Q_3 = mc_{\text{water}}\Delta T = 0.2 \times 4200 \times 100 = 84000 \text{ J}$

Stage 4: Vaporising water at $100°$C:

$Q_4 = mL_v = 0.2 \times 2.26 \times 10^6 = 452000 \text{ J}$

Stage 5: Steam from $100°$C to $110°$C:

$Q_5 = mc_{\text{steam}}\Delta T = 0.2 \times 2010 \times 10 = 4020 \text{ J}$

Total energy:

$Q_{\text{total}} = 6300 + 66800 + 84000 + 452000 + 4020 = 613120 \text{ J} = 613.1 \text{ kJ}$

(b) Total time:

$t = \frac{Q_{\text{total}}}{P} = \frac{613120}{500} = 1226.2 \text{ s} = 20.4 \text{ min}$

Key insight: The latent heat of vaporisation ($Q_4 = 452$ kJ) accounts for $73.7\%$ of the total energy. The phase change at $100°$C requires far more energy than raising the temperature of water by $100°$C ($Q_3 = 84$ kJ). This is the most common energy mistake in these problems.

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UT-2: Ideal Gas Law with Unit Traps

Question:

A gas cylinder contains $8$ g of oxygen ($O_2$, molar mass $= 32$ g mol$^{-1}$) at a pressure of $2.5 \times 10^5$ Pa and temperature $27°$C. (a) Calculate the volume of the cylinder. (b) If the temperature is raised to $127°$C while the volume is kept constant, what is the new pressure? (c) If the gas is then allowed to expand isothermally until the pressure returns to $2.5 \times 10^5$ Pa, what is the final volume?

Solution:

(a) Volume:

Critical step: Convert temperature to Kelvin: $T = 27 + 273 = 300$ K.

Number of moles: $n = \frac{m}{M} = \frac{8}{32} = 0.25$ mol

$PV = nRT$

$V = \frac{nRT}{P} = \frac{0.25 \times 8.31 \times 300}{2.5 \times 10^5}$

$V = \frac{623.25}{2.5 \times 10^5} = 2.493 \times 10^{-3} \text{ m}^3 = 2.49 \text{ L}$

(b) New pressure at constant volume:

$T_2 = 127 + 273 = 400$ K (must use Kelvin!)

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

$P_2 = P_1 \times \frac{T_2}{T_1} = 2.5 \times 10^5 \times \frac{400}{300} = 3.333 \times 10^5 \text{ Pa}$

Common mistake: Using $127/27$ instead of $400/300$ gives a completely wrong answer. Temperature in gas laws MUST always be in Kelvin.

(c) Final volume (isothermal expansion):

$T$ is constant at $400$ K:

$P_2 V_2 = P_3 V_3$

$3.333 \times 10^5 \times 2.493 \times 10^{-3} = 2.5 \times 10^5 \times V_3$

$V_3 = \frac{3.333 \times 2.493}{2.5} \times 10^{-3} = \frac{8.310}{2.5} \times 10^{-3} = 3.324 \times 10^{-3} \text{ m}^3 = 3.32 \text{ L}$

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UT-3: Kinetic Theory Pressure Derivation and Calculation

Question:

(a) Starting from the assumptions of the kinetic theory, derive the expression $p = \frac{1}{3}\rho\langle c^2 \rangle$ for the pressure exerted by an ideal gas. (b) Oxygen gas ($M = 32$ g mol$^{-1}$) is at temperature $300$ K and pressure $1.01 \times 10^5$ Pa. Calculate the RMS speed of the molecules and the density of the gas.

Solution:

(a) Derivation:

Consider $N$ molecules in a cubical container of side $L$, each of mass $m$.

A single molecule with velocity component $v_x$ in the $x$-direction travels between opposite walls in time $\Delta t = 2L / v_x$. Each collision with a wall reverses $v_x$, so the change in momentum per collision $= 2mv_x$.

Force on the wall from this molecule:

$F = \frac{\Delta p}{\Delta t} = \frac{2mv_x}{2L/v_x} = \frac{mv_x^2}{L}$

Total force on the wall from all molecules:

$F = \sum \frac{mv_{xi}^2}{L} = \frac{m}{L}\sum v_{xi}^2 = \frac{mN\langle v_x^2\rangle}{L}$

Since the motion is random: $\langle v_x^2\rangle = \langle v_y^2\rangle = \langle v_z^2\rangle = \frac{1}{3}\langle c^2\rangle$

$p = \frac{F}{L^2} = \frac{mN\langle v_x^2\rangle}{L^3} = \frac{mN}{V} \times \frac{1}{3}\langle c^2\rangle = \frac{1}{3}\frac{Nm}{V}\langle c^2\rangle = \frac{1}{3}\rho\langle c^2\rangle$

(b) RMS speed and density:

$pV = nRT = \frac{N}{N_A}RT$

$p = \frac{Nm\langle c^2\rangle}{3V} = \frac{\rho\langle c^2\rangle}{3}$

Also: $\frac{1}{2}m\langle c^2\rangle = \frac{3}{2}\frac{R}{N_A}T = \frac{3}{2}kT$

$c_{\text{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.31 \times 300}{0.032}} = \sqrt{\frac{7479}{0.032}} = \sqrt{233719} = 483.4 \text{ m s}^{-1}$

Density: $\rho = \frac{pM}{RT} = \frac{1.01 \times 10^5 \times 0.032}{8.31 \times 300} = \frac{3232}{2493} = 1.296 \text{ kg m}^{-3}$

Key insight: The RMS speed depends only on temperature and molar mass, NOT on pressure. At the same temperature, lighter molecules move faster. This is why hydrogen escapes from planetary atmospheres more easily than heavier gases.

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Integration Tests

IT-1: Thermodynamic Cycle Efficiency (with Energy and Work)

Question:

An ideal monatomic gas undergoes the following cycle:

• Process A$\to$B: Isothermal expansion at $T = 400$ K from $V_A = 2$ L to $V_B = 6$ L.
• Process B$\to$C: Isochoric (constant volume) cooling until pressure $P_C = P_A$.
• Process C$\to$A: Isobaric (constant pressure) compression back to state A.

The gas has $n = 0.5$ mol. Find (a) the pressure, volume, and temperature at each state, (b) the work done, heat exchanged, and internal energy change for each process, and (c) the efficiency of the cycle.

Solution:

State A: $T_A = 400$ K, $V_A = 2 \times 10^{-3}$ m$^3$

$P_A = \frac{nRT_A}{V_A} = \frac{0.5 \times 8.31 \times 400}{2 \times 10^{-3}} = \frac{1662}{0.002} = 8.31 \times 10^5 \text{ Pa}$

State B: $T_B = 400$ K (isothermal), $V_B = 6 \times 10^{-3}$ m$^3$

$P_B = \frac{nRT_B}{V_B} = \frac{0.5 \times 8.31 \times 400}{6 \times 10^{-3}} = \frac{1662}{0.006} = 2.77 \times 10^5 \text{ Pa}$

State C: $P_C = P_A = 8.31 \times 10^5$ Pa, $V_C = V_B = 6 \times 10^{-3}$ m$^3$ (isochoric)

$T_C = \frac{P_C V_C}{nR} = \frac{8.31 \times 10^5 \times 6 \times 10^{-3}}{0.5 \times 8.31} = \frac{4986}{4.155} = 1200 \text{ K}$

Process A$\to$B (isothermal expansion):

$\Delta U = 0$ (isothermal for ideal gas)

$W_{AB} = nRT \ln\frac{V_B}{V_A} = 0.5 \times 8.31 \times 400 \times \ln 3 = 1662 \times 1.0986 = 1826 \text{ J}$

$Q_{AB} = W_{AB} = 1826 \text{ J} \text{ (heat absorbed)}$

Process B$\to$C (isochoric heating):

$W_{BC} = 0$ (no volume change)

For monatomic ideal gas: $C_V = \frac{3}{2}R$

Since $T_C = 1200$ K $> T_B = 400$ K, the gas heats up during B$\to$C (pressure increases from $2.77 \times 10^5$ to $8.31 \times 10^5$ Pa).

$\Delta U_{BC} = nC_V(T_C - T_B) = 0.5 \times \frac{3}{2} \times 8.31 \times (1200 - 400) = 0.5 \times 12.465 \times 800 = 4986 \text{ J}$

$Q_{BC} = \Delta U_{BC} = 4986 \text{ J} \text{ (heat absorbed)}$

Process C$\to$A (isobaric compression):

$W_{CA} = P_C(V_A - V_C) = 8.31 \times 10^5 \times (2 - 6) \times 10^{-3} = -3324 \text{ J}$

$\Delta U_{CA} = nC_V(T_A - T_C) = 0.5 \times \frac{3}{2} \times 8.31 \times (400 - 1200) = 0.5 \times 12.465 \times (-800) = -4986 \text{ J}$

$Q_{CA} = \Delta U_{CA} + W_{CA} = -4986 + (-3324) = -8310 \text{ J} \text{ (heat released)}$

(c) Efficiency:

$W_{\text{net}} = W_{AB} + W_{BC} + W_{CA} = 1826 + 0 + (-3324) = -1498 \text{ J}$

The negative sign means net work is done ON the gas (this is a refrigeration cycle, not a heat engine). To get a heat engine, we should reverse the cycle direction. But proceeding with the given cycle:

$Q_{\text{absorbed}} = Q_{AB} + Q_{BC} = 1826 + 4986 = 6812 \text{ J}$

Since net work is negative (work done on gas), this is not a conventional heat engine cycle. The "efficiency" as a refrigerator would be:

$\text{COP} = \frac{Q_{\text{absorbed}}}{|W_{\text{net}}|} = \frac{6812}{1498} = 4.55$

Key insight: Not every thermodynamic cycle is a heat engine. The sign of the net work determines whether it operates as an engine (net work out) or a refrigerator (net work in). Students must check the signs carefully.

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IT-2: Adiabatic Process and Work Calculation (with Mechanics)

Question:

A monatomic ideal gas ($\gamma = 5/3$) is compressed adiabatically from $V_1 = 10$ L at $P_1 = 1 \times 10^5$ Pa to $V_2 = 4$ L. Find (a) the final pressure, (b) the final temperature (initial temperature $= 300$ K), (c) the work done on the gas, and (d) the change in internal energy.

Solution:

(a) Final pressure (adiabatic):

$P_1 V_1^\gamma = P_2 V_2^\gamma$

$P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma = 1 \times 10^5 \times \left(\frac{10}{4}\right)^{5/3} = 1 \times 10^5 \times 2.5^{1.667}$

$2.5^{1.667} = e^{1.667 \ln 2.5} = e^{1.667 \times 0.9163} = e^{1.5272} = 4.605$

$P_2 = 4.605 \times 10^5 \text{ Pa}$

(b) Final temperature:

$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$

$T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1} = 300 \times 2.5^{2/3} = 300 \times e^{0.6667 \times 0.9163} = 300 \times e^{0.6109} = 300 \times 1.842 = 552.6 \text{ K}$

(c) Work done on the gas:

For an adiabatic process, $Q = 0$, so $W = -\Delta U$.

$W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} = \frac{1 \times 10^5 \times 10 \times 10^{-3} - 4.605 \times 10^5 \times 4 \times 10^{-3}}{5/3 - 1}$

$W = \frac{1000 - 1842}{2/3} = \frac{-842}{0.6667} = -1263 \text{ J}$

Negative work means work is done ON the gas (compression). So work done on gas $= 1263$ J.

(d) Change in internal energy:

$\Delta U = nC_V(T_2 - T_1)$

For monatomic gas: $C_V = \frac{3}{2}R$

$n = \frac{P_1 V_1}{RT_1} = \frac{1 \times 10^5 \times 10 \times 10^{-3}}{8.31 \times 300} = \frac{1000}{2493} = 0.4011 \text{ mol}$

$\Delta U = 0.4011 \times \frac{3}{2} \times 8.31 \times (552.6 - 300) = 0.4011 \times 12.465 \times 252.6 = 1263 \text{ J}$

Check: $\Delta U = -W = 1263$ J (consistent, since $Q = 0$).

Key insight: In an adiabatic compression, ALL the work done on the gas increases its internal energy (and thus its temperature). This is why pumping air into a bicycle tyre makes the pump warm.

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IT-3: Mixing Liquids at Different Temperatures (with Nuclear Physics cross-topic)

Question:

$300$ g of water at $80°$C is mixed with $200$ g of water at $20°$C in an insulated calorimeter of heat capacity $50 \text{ J K}^{-1}$. Find the final equilibrium temperature. Then, a $100$ g aluminium block ($c_{\text{Al}} = 900 \text{ J kg}^{-1}\text{K}^{-1}$) at $150°$C is added to the mixture. Find the new equilibrium temperature.

Solution:

Stage 1: Mixing two water samples:

Let $T_f$ be the final temperature.

$m_1 c_w (T_1 - T_f) = m_2 c_w (T_f - T_2) + C_{\text{cal}}(T_f - T_2)$

$0.3 \times 4200 \times (80 - T_f) = 0.2 \times 4200 \times (T_f - 20) + 50(T_f - 20)$

$1260(80 - T_f) = 840(T_f - 20) + 50(T_f - 20)$

$100800 - 1260T_f = 890T_f - 17800$

$100800 + 17800 = 890T_f + 1260T_f$

$118600 = 2150T_f$

$T_f = 55.16°\text{C}$

Stage 2: Adding aluminium block:

Let $T_{f2}$ be the new equilibrium temperature.

$m_{\text{Al}} c_{\text{Al}} (150 - T_{f2}) = (m_1 + m_2) c_w (T_{f2} - 55.16) + C_{\text{cal}}(T_{f2} - 55.16)$

$0.1 \times 900 \times (150 - T_{f2}) = 0.5 \times 4200 \times (T_{f2} - 55.16) + 50(T_{f2} - 55.16)$

$90(150 - T_{f2}) = 2100(T_{f2} - 55.16) + 50(T_{f2} - 55.16)$

$13500 - 90T_{f2} = 2150T_{f2} - 118594$

$13500 + 118594 = 2150T_{f2} + 90T_{f2}$

$132094 = 2240T_{f2}$

$T_{f2} = 58.97°\text{C}$

Key insight: The calorimeter absorbs some heat, which is often forgotten. Also, when mixing substances, the final temperature must be between the initial temperatures of all components. If the calculation gives a temperature outside this range, there is an error.