Question

Gas in a cylinder-piston device expands from state 1 $(p_1, V_1, T_1)$ to state 2 $(p_2, V_2, T_2)$. The expansion process is polytropic, i.e., $pV^n = \text{constant}$, $n \neq 1$. Assuming the ideal gas behaviour, the expression for the work done, $W$, by the system is given by:

• A. $W = p_1 V_1 \ln\!\left(\dfrac{T_2}{T_1}\right)$
• B. $W = \dfrac{p_2 V_2 - p_1 V_1}{1 - n}$
• C. $W = p_1 V_1 \ln\!\left(\dfrac{V_1}{V_2}\right)$
• D. $W = p_2 V_2 \ln\!\left(\dfrac{p_2}{p_1}\right)$

Hint:

For polytropic processes ($pV^n=\text{const}$), use $W=\dfrac{p_2V_2 - p_1V_1}{1-n}$ when $n\neq1$. Only for $n=1$ (isothermal) does the logarithmic form apply.

Answer 1

The Correct Option is B

For a polytropic process, the relation $pV^n = C$ (constant) holds. The work done by the gas during a quasi-static expansion from $V_1$ to $V_2$ is:
\[ W = \int_{V_1}^{V_2} p\, dV \] Using the polytropic relation, the pressure can be written as:
\[ p = \frac{C}{V^n} \] Thus, the work integral becomes:
\[ W = \int_{V_1}^{V_2} \frac{C}{V^n}\, dV \] Evaluating the integral for $n \neq 1$:
\[ W = \frac{C}{1-n}\left( V_2^{\,1-n} - V_1^{\,1-n} \right) \] Using $C = p_1 V_1^n = p_2 V_2^n$, the expression becomes:
\[ W = \frac{p_2 V_2 - p_1 V_1}{1-n} \] Thus, the correct expression for polytropic work is:
\[ W = \frac{p_2 V_2 - p_1 V_1}{1-n} \]