Question

One mole of a perfect gas expands isothermally and reversibly from \(10\text{dm}^3\) to \(20\text{dm}^3\) at \(300\text{ K}\) . Find \(\Delta \text{U}\), q and work done respectively in the process. \(\left( \text{R} = 8.3 \times 10^{-3} \text{ kJ K}^{-1} \text{ mol}^{-1} \right)\)

• A. \(0.0\text{ kJ}, 1.726\text{ kJ}, -1.726\text{ kJ}\)
• B. \(0.0\text{ kJ}, -17.48\text{ kJ}, -17.48\text{ kJ}\)
• C. \(-1.726\text{ kJ}, 0\text{ kJ}, 1.726\text{ kJ}\)
• D. \(0.0\text{ kJ}, 21.84\text{ kJ}, -21.84\text{ kJ}\)

Hint:

Isothermal rule: \(\Delta U = 0\), so \(q = -w\).

Answer 1

The Correct Option is A

Concept:
For isothermal process of an ideal gas: \[ \Delta U = 0 \] Work done: \[ w = -nRT \ln \frac{V_2}{V_1} \] Heat: \[ q = -w \]

Step 1: Calculate work done. \[ w = - (1)(8.3 \times 10^{-3})(300) \ln \left(\frac{20}{10}\right) \] \[ w = - (2.49) \ln(2) \] \[ \ln(2) \approx 0.693 \] \[ w = - (2.49 \times 0.693) = -1.726 \text{ kJ} \]

Step 2: Calculate heat. \[ q = -w = +1.726 \text{ kJ} \]

Step 3: Internal energy change. \[ \Delta U = 0 \]

Step 4: Conclusion. \[ \Delta U = 0,\quad q = +1.726\text{ kJ},\quad w = -1.726\text{ kJ} \]