Question

If \(2\) mole of an ideal gas expand isothermally and reversibly at \(27^{\circ}\)C from 1 \(\text{dm}^3\) to \(1 \text{m}^3\) calculate work done? \([R = 8.314 \text{J K}^{-1}\text{mol}^{-1}]\)

• A. \(-49.95\) kJ
• B. \(-99.90\) kJ
• C. \(-34.46\) kJ
• D. \(-68.92\) kJ

Hint:

Use W = -2.303 nRT log(V2/V1) with both volumes in the same unit.

Answer 1

The Correct Option is C

Step 1: Setup:
For an ideal gas at constant temperature the reversible work is $W = -nRT\ln(V_2/V_1)$. I will use the natural log directly.

Step 2: Numbers:
$nRT = 2 \times 8.314 \times 300 = 4988.4$ J.
Volume ratio $= 1\ \text{m}^3 / 1\ \text{dm}^3 = 10^3$, so $\ln 1000 = 3 \times 2.303 = 6.908$.

Step 3: Multiply:
\[ W = -4988.4 \times 6.908 = -34461 \text{ J} \approx -34.46 \text{ kJ} \]

Step 4: Check:
The sign is negative because the gas expands and does work. A 1000-fold expansion at 300 K for 2 mol should give a few tens of kJ, which agrees.

Final Answer:
The work done is -34.46 kJ, option (C). \[ \boxed{-34.46 \text{ kJ}} \]