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: Understanding the Concept:
In a reversible isothermal expansion of an ideal gas the system does work on the surroundings, so the work done by the gas appears with a negative sign in the convention used for MHT CET.

Step 2: Key Formula:
\[ W = -2.303\, nRT \log_{10}\frac{V_2}{V_1} \]

Step 3: Detailed Explanation:
Convert volumes to the same unit: \(1\ \text{m}^3 = 1000\ \text{dm}^3\), so \(V_2/V_1 = 1000/1 = 1000\) and \(\log_{10}1000 = 3\).
Temperature \(T = 27 + 273 = 300\) K and \(n = 2\).
\[ W = -2.303 \times 2 \times 8.314 \times 300 \times 3 \]
\[ 2.303 \times 2 \times 8.314 = 38.29,\quad 38.29 \times 300 \times 3 = 34461 \text{ J} \]
\[ W \approx -34.46 \text{ kJ} \]

Step 4: Why the other options are wrong.
Option (A) \(-49.95\) kJ and (B) \(-99.90\) kJ come from using a wrong volume ratio or a wrong log value. Option (D) \(-68.92\) kJ is exactly double the correct value, so it results from using \(n = 4\) or an extra factor of 2.

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