Heat and Thermodynamics

Concept:The processes involve an adiabatic compression followed by isochoric cooling and isothermal expansion.Explanation:The sudden compression from state $a$ to $b$ is adiabatic because there is no time for heat exchange.For an adiabatic process, $T V^{\gamma-1}$ is constant.For a monoatomic ideal gas, $\gamma = \frac{5}{3}$ so $\gamma-1 = \frac{2}{3}$.Given $T_a = 27^\circ \mathrm{C} = 300\,\mathrm{K}$ and $V_a = V_0$, $V_b = V_0/3$.From $T_a V_a^{\gamma-1} = T_b V_b^{\gamma-1}$, we get $300 \times V_0^{2/3} = T_b \times (V_0/3)^{2/3}$.Thus $T_b = 300 \times 3^{2/3} = 300 \times (9)^{1/3}$.Using $(9)^{1/3} = 2.08$, $T_b \approx 624\,\mathrm{K}$.Pressure at $b$: $P_b = P_a \left(\frac{V_a}{V_b}\right)^\gamma = P_a \times 3^{5/3} \approx 6.24\,P_a$, not $2.08\,P_a$. So option D is wrong.Change in internal energy from $a$ to $b$: $\Delta U_{a\to b} = n C_v (T_b - T_a)$.$n = 10$, $C_v = \frac{3}{2}R$, so $\Delta U = 10 \times \frac{3}{2}R \times (624-300) = 15R \times 324 = 4860R$. Option B is correct.Process $b \to c$ is isochoric (constant volume) while gas cools to water bath temperature $T_c = 11^\circ\mathrm{C} = 284\,\mathrm{K}$.Process $c \to f$ is isothermal at $284\,\mathrm{K}$; piston is brought back to original volume $V_f = V_0$.The schematic P‑V diagram correctly shows: adiabatic steep curve ($a\to b$), vertical isochoric line ($b\to c$), isothermal curve ($c\to f$).Thus option A is correct.Net change in internal energy for the whole cycle depends only on initial and final temperatures (state function): $\Delta U_{\text{net}} = n C_v (T_f - T_a)$.$T_a = 300\,\mathrm{K}$, $T_f = 284\,\mathrm{K}$, so $\Delta U_{\text{net}} = 10 \times \frac{3}{2}R \times (284-300) = 15R \times (-16) = -240R$. Option C is correct.Answer:Options A, B, C.