Thermodynamics Final Exam
- Student ID: 1W23CF13
- Name: TAI Yungche
- Date: 2025/07/22
Atmosphere of a planet X
Question 1
The molar mass of air at ground level is calculated using the weighted average based on the number fractions of each component.
Given:
- \(r_{CO_2}^0 = 0.70\) (number fraction of CO₂)
- \(r_{He}^0 = 0.30\) (number fraction of He)
- \(\mu_{CO_2} = 44 \text{ g/mol}\)
- \(\mu_{He} = 4 \text{ g/mol}\)
The molar mass of the air mixture: \(\(\mu_0 = r_{CO_2}^0 \cdot \mu_{CO_2} + r_{He}^0 \cdot \mu_{He}\)\)
Question 2
For an isothermal atmosphere, the pressure variation with altitude follows the barometric formula: \(\(p(h) = p_0 \exp\left(-\frac{\mu g h}{RT}\right)\)\)
Since each gas component behaves independently in an ideal gas mixture, the partial pressures follow: \(\(p_{CO_2}(h) = p_{CO_2}^0 \exp\left(-\frac{\mu_{CO_2} g h}{RT}\right)\)\) \(\(p_{He}(h) = p_{He}^0 \exp\left(-\frac{\mu_{He} g h}{RT}\right)\)\)
At ground level:
- \(p_{CO_2}^0 = r_{CO_2}^0 \cdot p_0 = 0.70 p_0\)
- \(p_{He}^0 = r_{He}^0 \cdot p_0 = 0.30 p_0\)
The number fractions at height \(h\) are: \(\(r_{CO_2}^h = \frac{p_{CO_2}(h)}{p_{CO_2}(h) + p_{He}(h)}\)\)
Substituting the exponential expressions: \(\(r_{CO_2}^h = \frac{0.70 \exp\left(-\frac{\mu_{CO_2} g h}{RT}\right)}{0.70 \exp\left(-\frac{\mu_{CO_2} g h}{RT}\right) + 0.30 \exp\left(-\frac{\mu_{He} g h}{RT}\right)}\)\)
Given values:
- \(h = 10000 \text{ m}\)
- \(g = 20 \text{ m/s}^2\)
- \(T = 27°C = 300 \text{ K}\)
- \(R = 8.314 \text{ J/(mol·K)}\)
Calculate the exponents: \(\(\frac{\mu_{CO_2} g h}{RT} = \frac{44 \times 10^{-3} \times 20 \times 10000}{8.314 \times 300} = \frac{8800}{2494.2} = 3.529\)\)
The exponential terms: \(\(\exp(-3.529) = 0.0294\)\) \(\(\exp(-0.321) = 0.726\)\)
Therefore: \(\(r_{CO_2}^h = \frac{0.70 \times 0.0294}{0.70 \times 0.0294 + 0.30 \times 0.726} = \frac{0.0206}{0.0206 + 0.218} = \frac{0.0206}{0.239} = 0.086\)\)
The molar mass at height \(h\): \(\(\mu_h = r_{CO_2}^h \cdot \mu_{CO_2} + r_{He}^h \cdot \mu_{He}\)\)
The air composition at the top of the tallest mountain:
- CO₂: 8.6%
- He: 91.4%
- Molar mass: 7.44 g/mol
Question 3
For CO₂ at T = 27°C = 300 K, the molecule has linear geometry with:
- 3 translational degrees of freedom
- 2 rotational degrees of freedom (linear molecule)
- 4 vibrational modes: (3N - 5) = (3×3 - 5) = 4
Using the equipartition theorem:
- Each translational and rotational degree contributes R/2 to \(c_V\)
- Each vibrational mode contributes R when fully excited
For CO₂: \(\(c_V^{CO_2} = \frac{3R}{2} + \frac{2R}{2} + n_{vib} \cdot R\)\)
At 300 K, not all vibrational modes are fully excited. The vibrational contribution is negligible, so: \(\(c_V^{CO_2} = \frac{5R}{2} = \frac{5 \times 8.314}{2} = 20.785 \text{ J/(mol·K)}\)\)
For the air mixture at ground level: \(\(c_V^{air} = r_{CO_2}^0 \cdot c_V^{CO_2} + r_{He}^0 \cdot c_V^{He}\)\) \(\(c_V^{air} = 0.70 \times 20.785 + 0.30 \times 12.471 = 14.550 + 3.741 = 18.291 \text{ J/(mol·K)}\)\)
Question 4
The adiabatic expansion coefficient γ is defined as: \(\(\gamma = \frac{c_P}{c_V} = \frac{c_V + R}{c_V}\)\)
At ground level (h = 0): \(\(c_V^0 = 18.291 \text{ J/(mol·K)}\)\) \(\(c_P^0 = c_V^0 + R = 18.291 + 8.314 = 26.605 \text{ J/(mol·K)}\)\) \(\(\gamma^0 = \frac{26.605}{18.291} = 1.455\)\)
At the tallest mountain (h = 10000 m): \(\(c_V^h = r_{CO_2}^h \cdot c_V^{CO_2} + r_{He}^h \cdot c_V^{He}\)\) \(\(c_V^h = 0.086 \times 20.785 + 0.914 \times 12.471 = 1.787 + 11.399 = 13.186 \text{ J/(mol·K)}\)\) \(\(c_P^h = c_V^h + R = 13.186 + 8.314 = 21.500 \text{ J/(mol·K)}\)\) \(\(\gamma^h = \frac{21.500}{13.186} = 1.631\)\)
The adiabatic expansion coefficients:
- At ground level: γ⁰ = 1.455
- At mountain top: γʰ = 1.631
Heat engine
Looking at this heat engine problem with an ideal gas undergoing a 1→2→3→1 cycle.
Question 1
For process 1→2 (isobaric expansion):
Since pressure is constant: \(P_1 = P_2\)
Work done: \(\(W_1 = \int_{V_1}^{V_2} P \, dV = P(V_2 - V_1)\)\)
Using the ideal gas law \(PV = Nk_BT\): \(\(W_1 = Nk_B(T_2 - T_1)\)\)
Internal energy change for ideal gas: \(\(\Delta U_1 = C_V(T_2 - T_1)\)\)
From the first law of thermodynamics: \(\(Q_1 = \Delta U_1 + W_1 = C_V(T_2 - T_1) + Nk_B(T_2 - T_1)\)\) \(\(Q_1 = (C_V + Nk_B)(T_2 - T_1)\)\)
Question 2
For process 2→3 (isochoric cooling):
Since volume is constant: \(V_2 = V_3\)
Work done: \(\(W_2 = 0\)\)
Internal energy change: \(\(\Delta U_2 = C_V(T_1 - T_2) = -C_V(T_2 - T_1)\)\)
Heat transferred: \(\(Q_2 = \Delta U_2 + W_2 = -C_V(T_2 - T_1)\)\)
Question 3
For process 3→1 (isothermal compression):
Since temperature is constant: \(T_3 = T_1\)
Internal energy change: \(\(\Delta U_3 = 0\)\)
Work done: \(\(W_3 = \int_{V_3}^{V_1} P \, dV = \int_{V_3}^{V_1} \frac{Nk_BT_1}{V} \, dV = Nk_BT_1 \ln\left(\frac{V_1}{V_3}\right)\)\)
From the isobaric process 1→2: \(\frac{V_1}{V_2} = \frac{T_1}{T_2}\)
Since \(V_3 = V_2\): \(\(W_3 = Nk_BT_1 \ln\left(\frac{T_1}{T_2}\right)\)\)
Heat transferred: \(\(Q_3 = W_3 = Nk_BT_1 \ln\left(\frac{T_1}{T_2}\right)\)\)
Question 4
Net work done by the gas in the cycle: \(\(W = W_1 + W_2 + W_3\)\) \(\(W = Nk_B(T_2 - T_1) + 0 + Nk_BT_1 \ln\left(\frac{T_1}{T_2}\right)\)\) \(\(W = Nk_B\left[(T_2 - T_1) + T_1 \ln\left(\frac{T_1}{T_2}\right)\right]\)\)
Question 5
Total heat absorbed by the gas occurs only in process 1→2 where \(Q_1 > 0\).
In processes 2→3 and 3→1, heat is released (\(Q_2 < 0\) and \(Q_3 < 0\) since \(T_1 < T_2\)).
Therefore, total heat absorbed: \(\(Q_{+} = Q_1 = (C_V + Nk_B)(T_2 - T_1)\)\)
Question 6
Total heat released by the gas occurs in processes where heat flows out of the system.
Process 2→3: \(Q_2 = -C_V(T_2 - T_1) < 0\) (heat released)
Process 3→1: \(Q_3 = Nk_BT_1 \ln\left(\frac{T_1}{T_2}\right) < 0\) (heat released since \(T_1 < T_2\))
Total heat released: \(\(Q_{-} = |Q_2| + |Q_3| = C_V(T_2 - T_1) + Nk_BT_1 \ln\left(\frac{T_2}{T_1}\right)\)\)
Question 7
Efficiency of the cycle: \(\(\eta = \frac{W}{Q_{+}} = \frac{W}{Q_1}\)\)
From Question 4: \(W = Nk_B\left[(T_2 - T_1) + T_1 \ln\left(\frac{T_1}{T_2}\right)\right]\)
From Question 5: \(Q_1 = (C_V + Nk_B)(T_2 - T_1)\)
Therefore: \(\(\eta = \frac{Nk_B\left[(T_2 - T_1) + T_1 \ln\left(\frac{T_1}{T_2}\right)\right]}{(C_V + Nk_B)(T_2 - T_1)}\)\)
For \(T_2 = 2T_1\): \(\(\eta = \frac{Nk_B\left[T_1 + T_1 \ln\left(\frac{1}{2}\right)\right]}{(C_V + Nk_B)T_1} = \frac{Nk_B(1 - \ln 2)}{C_V + Nk_B}\)\)
The Carnot efficiency between \(T_1\) and \(T_2 = 2T_1\): \(\(\eta_{Carnot} = 1 - \frac{T_1}{T_2} = 1 - \frac{1}{2} = 0.5\)\)
Since \((1 - \ln 2) \approx 0.307 < 0.5\) and \(\frac{Nk_B}{C_V + Nk_B} < 1\), we have: \(\(\eta < \eta_{Carnot}\)\)
Question 8
Entropy change for process 1→2 (isobaric): \(\(\Delta_1 S = \int_{T_1}^{T_2} \frac{C_P}{T} dT = C_P \ln\left(\frac{T_2}{T_1}\right)\)\)
where \(C_P = C_V + Nk_B\) for ideal gas.
Entropy change for process 2→3 (isochoric): \(\(\Delta_2 S = \int_{T_2}^{T_1} \frac{C_V}{T} dT = C_V \ln\left(\frac{T_1}{T_2}\right)\)\)
Entropy change for process 3→1 (isothermal): \(\(\Delta_3 S = \int \frac{dQ}{T} = \frac{Q_3}{T_1} = \frac{W_3}{T_1} = Nk_B \ln\left(\frac{T_1}{T_2}\right)\)\)
Verification: Total entropy change for the complete cycle should be zero: \(\(\Delta S_{total} = \Delta_1 S + \Delta_2 S + \Delta_3 S = (C_V + Nk_B) \ln\left(\frac{T_2}{T_1}\right) + C_V \ln\left(\frac{T_1}{T_2}\right) + Nk_B \ln\left(\frac{T_1}{T_2}\right) = 0\)\)