Question

An ideal gas undergoes an isothermal process at 300 K. If the initial volume is 2 m\(^3\) and it doubles to 4 m\(^3\), what is the work done by the gas? (Take \(R = 8.31 \, \text{J/(mol.K)}\), and assume 1 mole of gas.)

• A. 1728 J
• B. 2076 J
• C. 2492 J
• D. 3000 J

Hint:

For work done in an isothermal process, use \( W = nRT \ln\left(\frac{V_f}{V_i}\right) \), with \( R = 8.31 \, \text{J/(mol·K)} \).

Answer 1

The Correct Option is A

For an isothermal process, the work done by an ideal gas is: \[ W = nRT \ln\left(\frac{V_f}{V_i}\right) \] where \( n = 1 \, \text{mol} \),\(R = 8.31 \, \text{J/(mol.K)}\), \( T = 300 \, \text{K} \), \( V_i = 2 \, \text{m}^3 \), and \( V_f = 4 \, \text{m}^3 \). Calculate: \[ \frac{V_f}{V_i} = \frac{4}{2} = 2 \implies \ln(2) \approx 0.693 \] \[ W = 1 \cdot 8.31 \cdot 300 \cdot 0.693 = 2493 \cdot 0.693 \approx 1727.6 \, \text{J} \] \[ {1728} \] (Note: The calculation yields ~1728 J; option (B) may assume a different constant or condition.)