Question

1 mole of an ideal gas is compressed isothermally and reversibly from initial pressure \(x\) kPa to final pressure \(2x\) kPa at 300 K. Find the work done \((R = 8.314 \text{J K}^{-1}\text{mol}^{-1})\)

• A. \(1432\) J
• B. \(1865\) J
• C. \(1296\) J
• D. \(1729\) J

Hint:

Use w = nRT ln(P2/P1) for isothermal reversible compression with P2/P1 = 2.

Answer 1

The Correct Option is D

Step 1: Natural log route:
$W = nRT\ln 2$ with $\ln 2 = 0.6931$.
$nRT = 1 \times 8.314 \times 300 = 2494.2$ J.

Step 2: Multiply:
$W = 2494.2 \times 0.6931 = 1728.8 \approx 1729$ J.

Step 3: Interpretation:
Doubling the pressure at constant temperature halves the volume, so $V_2/V_1 = 1/2$. Then $-nRT\ln(V_2/V_1) = +nRT\ln 2$. The value is positive since the surroundings do work on the gas.

Step 4: Check the alternatives:
Each distractor differs from 1729 J by an arithmetic slip. For instance, $2494.2 \times 0.5 = 1247$ J, which is not listed, so only (D) fits.

Final Answer:
$W \approx 1729$ J, option (D). \[ \boxed{1729 \text{ J (D)}} \]