Question

Two samples 1 and 2 are initially kept in the same state . The sample 1 is expanded through isothermal process where as sample 2 through adiabatic process upto the same final volume . let w1 and w2 be the work done by the systems 1 and 2 respectively then :

• A. w1<w2
• B. w1>w2
• C. w1=w2
• D. None

Answer 1

Answer: (B)

w1>w2

Answer 2

Let the initial state of both samples be $(P_1, V_1)$. Both samples expand to the same final volume $V_2$, where $V_2 > V_1$.

For sample 1, the process is isothermal. The equation of state is $PV = \text{constant}$. Let the final state be $(P_{2,iso}, V_2)$. The work done by the system during isothermal expansion is given by: $W_1 = \int_{V_1}^{V_2} P dV$ Since $PV = P_1 V_1$, we have $P = \frac{P_1 V_1}{V}$. $W_1 = \int_{V_1}^{V_2} \frac{P_1 V_1}{V} dV = P_1 V_1 \int_{V_1}^{V_2} \frac{1}{V} dV = P_1 V_1 [\ln V]_{V_1}^{V_2} = P_1 V_1 \ln\left(\frac{V_2}{V_1}\right)$. Alternatively, using the ideal gas law $P_1 V_1 = nRT_1$ (where $T_1$ is the constant temperature), $W_1 = nRT_1 \ln\left(\frac{V_2}{V_1}\right)$. Since $V_2 > V_1$, $\ln(V_2/V_1) > 0$, so $W_1 > 0$.

For sample 2, the process is adiabatic. The equation of state is $PV^\gamma = \text{constant}$, where $\gamma = C_p/C_v > 1$. Let the final state be $(P_{2,adi}, V_2)$. The work done by the system during adiabatic expansion is given by: $W_2 = \int_{V_1}^{V_2} P dV$ Since $PV^\gamma = P_1 V_1^\gamma$, we have $P = \frac{P_1 V_1^\gamma}{V^\gamma}$. $W_2 = \int_{V_1}^{V_2} \frac{P_1 V_1^\gamma}{V^\gamma} dV = P_1 V_1^\gamma \int_{V_1}^{V_2} V^{-\gamma} dV = P_1 V_1^\gamma \left[\frac{V^{-\gamma+1}}{-\gamma+1}\right]_{V_1}^{V_2} = \frac{P_1 V_1^\gamma}{1-\gamma} (V_2^{1-\gamma} - V_1^{1-\gamma})$. $W_2 = \frac{P_1 V_1^\gamma}{1-\gamma} V_1^{1-\gamma} \left(\left(\frac{V_2}{V_1}\right)^{1-\gamma} - 1\right) = \frac{P_1 V_1}{1-\gamma} \left(\left(\frac{V_2}{V_1}\right)^{1-\gamma} - 1\right)$. Since $\gamma > 1$, $1-\gamma < 0$. Let $1-\gamma = -\alpha$ where $\alpha = \gamma-1 > 0$. $W_2 = \frac{P_1 V_1}{-\alpha} \left(\left(\frac{V_2}{V_1}\right)^{-\alpha} - 1\right) = \frac{P_1 V_1}{\alpha} \left(1 - \left(\frac{V_1}{V_2}\right)^{\alpha}\right) = \frac{P_1 V_1}{\gamma-1} \left(1 - \left(\frac{V_1}{V_2}\right)^{\gamma-1}\right)$. Since $V_2 > V_1$, $V_1/V_2 < 1$, so $(V_1/V_2)^{\gamma-1} < 1$. Thus, $1 - (V_1/V_2)^{\gamma-1} > 0$, and $W_2 > 0$.

To compare $W_1$ and $W_2$, we can consider the P-V diagram. Both processes start at the same point $(P_1, V_1)$ and expand to the same final volume $V_2$. The work done is the area under the curve on the P-V diagram. For $V > V_1$, the pressure in the isothermal process is $P_{iso}(V) = P_1 \frac{V_1}{V}$. For $V > V_1$, the pressure in the adiabatic process is $P_{adi}(V) = P_1 \left(\frac{V_1}{V}\right)^\gamma$. Since $V_2 > V \ge V_1$ and $\gamma > 1$, we have $\frac{V_1}{V} \le 1$ and $\left(\frac{V_1}{V}\right)^\gamma \le \frac{V_1}{V}$. Thus, $P_{adi}(V) \le P_{iso}(V)$ for $V \ge V_1$. For $V > V_1$, $P_{adi}(V) < P_{iso}(V)$. This means the isothermal curve lies above the adiabatic curve for $V > V_1$. Since the work done is the area under the curve $\int_{V_1}^{V_2} P dV$, and $P_{iso}(V) > P_{adi}(V)$ for $V_1 < V \le V_2$, the area under the isothermal curve is greater than the area under the adiabatic curve. Therefore, $W_1 = W_{iso} > W_{adi} = W_2$.

The relationship between $W_1$ and $W_2$ is $W_1 > W_2$.

Explanation:

Work done by a system during expansion is the area under the pressure-volume (P-V) curve. For expansion from the same initial state to the same final volume, the isothermal process curve ($PV = \text{constant}$) lies above the adiabatic process curve ($PV^\gamma = \text{constant}$, where $\gamma > 1$) on the P-V diagram for $V > V_1$. This is because pressure drops faster in an adiabatic expansion than in an isothermal expansion. Consequently, the area under the isothermal curve ($W_1$) is greater than the area under the adiabatic curve ($W_2$). Therefore, $W_1 > W_2$.