Question

\(20.0\,\text{dm}^3\) of an ideal gas \(X\) at \(600\) K and \(0.5\) MPa undergoes isothermal reversible expansion until the pressure of the gas becomes \(0.2\) MPa. Which of the following option is correct? (Given: \(\log 2 = 0.3010\), \(\log 5 = 0.6989\))

• A. \(w=-3.9\,\text{kJ},\ \Delta U=0,\ \Delta H=0,\ q=3.9\,\text{kJ}\)
• B. \(w=9.1\,\text{kJ},\ \Delta U=9.1\,\text{kJ},\ \Delta H=0,\ q=0\)
• C. \(w=-9.1\,\text{kJ},\ \Delta U=0,\ \Delta H=0,\ q=9.1\,\text{kJ}\)
• D. \(w=+4.1\,\text{kJ},\ \Delta U=0,\ \Delta H=0,\ q=-4.1\,\text{kJ}\)

Hint:

In isothermal processes of ideal gases, work done is exactly balanced by heat absorbed or released.

Answer 1

The Correct Option is C

Concept: For an {ideal gas undergoing isothermal process}: \[ \Delta U = 0,\quad \Delta H = 0 \] Work done in reversible isothermal expansion: \[ w = -nRT \ln\!\left(\frac{V_2}{V_1}\right) = -nRT \ln\!\left(\frac{P_1}{P_2}\right) \] Heat absorbed: \[ q = -w \]
Step 1: Convert logarithm \[ \frac{P_1}{P_2}=\frac{0.5}{0.2}=2.5 \] \[ \ln(2.5)=2.303(\log 5 - \log 2) =2.303(0.6989-0.3010) =0.916 \]
Step 2: Calculate number of moles \[ n=\frac{PV}{RT} =\frac{0.5\times10^6 \times 20\times10^{-3}} {8.314\times600} \approx 2.0 \]
Step 3: Calculate work done \[ w = -nRT\ln\left(\frac{P_1}{P_2}\right) = -2 \times 8.314 \times 600 \times 0.916 \approx -9.1\,\text{kJ} \]
Step 4: Determine heat and energy changes \[ \Delta U=0,\quad \Delta H=0,\quad q=+9.1\,\text{kJ} \] Final Answer: \[ \boxed{w=-9.1\,\text{kJ},\ \Delta U=0,\ \Delta H=0,\ q=9.1\,\text{kJ}} \]