Degrees of Freedom

Thermodynamic process equations govern individual state paths: sudden compression $$a \rightarrow b$$ is adiabatic, stationary cooling $$b \rightarrow c$$ is isochoric, and slow expansion $$c \rightarrow f$$ inside the water bath is isothermal.

Given: $$n = 10$$, $$\gamma = \frac{5}{3}$$, $$C_v = \frac{3}{2}R$$, $$T_a = 27^\circ\text{C} = 300\text{ K}$$, $$T_c = T_f = 11^\circ\text{C} = 284\text{ K}$$, $$V_a = V_f = V_0$$, $$V_b = V_c = \frac{V_0}{3}$$, $$9^{1/3} = 2.08$$

Along adiabatic path $$a \rightarrow b$$:

$$T_a V_a^{\gamma-1} = T_b V_b^{\gamma-1}$$

$$300 \times V_0^{2/3} = T_b \left(\frac{V_0}{3}\right)^{2/3} \implies T_b = 300 \times 3^{2/3} = 300 \times 9^{1/3} = 300 \times 2.08 = 624\text{ K}$$

$$P_a V_a^\gamma = P_b V_b^\gamma \implies P_b = P_a\left(\frac{V_a}{V_b}\right)^\gamma = P_a(3)^{5/3} = P_a \times 3 \times 3^{-1/3} \approx 6.24 P_a$$

$$\Delta U_{a \rightarrow b} = nC_v(T_b - T_a) = 10 \times \frac{3}{2}R \times (624 - 300) = 15R \times 324 = 4860R$$

Evaluating net change in internal energy for the entire process $$a \rightarrow b \rightarrow c \rightarrow f$$:

$$\Delta U_{\text{net}} = nC_v(T_f - T_a) = 10 \times \frac{3}{2}R \times (284 - 300) = 15R \times (-16) = -240R$$

Answer: Option (A), Option (B), Option (C)

Given: Monoatomic gas $$\implies \gamma = \frac{5}{3}$$, $$C_v = \frac{3}{2}R$$, $$\ln 2 \approx 0.7$$

Along adiabatic compression $$c \rightarrow a$$:

$$T_c V_c^{\gamma-1} = T_a V_a^{\gamma-1} \implies \frac{T_a}{T_c} = \left(\frac{V_c}{V_a}\right)^{\gamma-1} = \left(\frac{V_2}{V_1}\right)^{2/3}$$

At $$\frac{V_2}{V_1} = 8$$:

$$\frac{T_a}{T_c} = (8)^{2/3} = 4 \implies T_a = 4T_c$$

Along isothermal expansion $$a \rightarrow b$$:

$$T_a = T_b \implies P_a V_1 = P_b V_2 \implies \frac{P_a}{P_b} = \frac{V_2}{V_1} = 8$$

$$Q_{\text{in}} = Q_{ab} = nRT_a \ln\left(\frac{V_2}{V_1}\right) = nRT_a \ln 8 = 3nRT_a \ln 2 \approx 2.1nRT_a$$

Along isochoric cooling $$b \rightarrow c$$:

$$Q_{\text{out}} = \vert{}Q_{bc}\vert{} = nC_v(T_b - T_c) = \frac{3}{2}nR(T_a - T_c) = \frac{3}{2}nR\left(T_a - \frac{T_a}{4}\right) = \frac{9}{8}nRT_a = 1.125nRT_a$$

$$Q_{\text{out}} < Q_{\text{in}}$$

Evaluating efficiency $$\eta$$:

$$\eta = \frac{Q_{\text{in}} - Q_{\text{out}}}{Q_{\text{in}}} = 1 - \frac{\frac{9}{8}nRT_a}{3nRT_a\ln 2} = 1 - \frac{3}{8\ln 2}$$

Using heat exchange relations:

$$dQ = -C_{v1} dT_1 = C_{p2} dT_2$$

$$dT_1 = -\frac{2 dQ}{3R},\quad dT_2 = \frac{2 dQ}{5R}$$

$$d(\Delta T) = dT_1 - dT_2 = -dQ\left(\frac{2}{3R} + \frac{2}{5R}\right) = -\frac{16}{15R} dQ$$

Substituting $$dQ = \frac{KA}{x}\Delta T dt$$:

$$\frac{d(\Delta T)}{\Delta T} = -\frac{16KA}{15xR} dt$$

$$\int_{\Delta T_0}^{\Delta T_0/2} \frac{d(\Delta T)}{\Delta T} = -\frac{16KA}{15xR} \int_{0}^{t} dt$$

$$\ln 2 = \frac{16KA}{15xR} t \implies t = \frac{15\ln 2}{16}\frac{xR}{KA}$$

$$n = \frac{15 \times 0.7}{16} = 0.65625 \approx 0.66$$