Thermodynamics Part 5

Concept:For an ideal gas, isothermal work depends on $\ln(V_2/V_1)$ and adiabatic work depends on temperature change using $\gamma$.Explanation:Given: isothermal expansion from $V$ to $2V$ at $27^\circ\text{C}$ ($T=300\,\text{K}$).Work done: $W = nRT \ln\frac{V_2}{V_1} = 1 \times R \times 300 \times \ln 2$.Since $\ln 2 = 0.693$, $W = 300R \times 0.693 = 207.9R$.Now adiabatic expansion does same work $W=207.9R$, same initial temperature $T_1=300\,\text{K}$, same amount of gas $n=1$.For diatomic gas, $\gamma = \frac{7}{5} = 1.4$, so $\gamma-1 = 0.4$.Adiabatic work: $W = \frac{nR(T_1 - T_2)}{\gamma-1} = \frac{R(300 - T_2)}{0.4}$.Equate: $207.9R = \frac{R(300 - T_2)}{0.4}$. Cancel $R$: $207.9 = \frac{300 - T_2}{0.4}$.Multiply: $207.9 \times 0.4 = 300 - T_2$ → $83.16 = 300 - T_2$.Thus $T_2 = 300 - 83.16 = 216.84\,\text{K}$.Convert to Celsius: $T_2(^\circ\text{C}) = 216.84 - 273.15 = -56.31^\circ\text{C} \approx -56^\circ\text{C}$.Answer:Option A: $-56^\circ\text{C}$.