Thermodynamics — Easy JEE chemistry MCQ
Two moles of an ideal gas undergo a reversible isothermal expansion at 300 K, where the volume increases from 10 L to 20 L. The work done by the gas is (R = 8.314 J/mol·K, ln 2 = 0.693):
- A. 1.73 kJ
- B. 3.46 kJ
- C. 6.92 kJ
- D. 13.84 kJ
Solution
For reversible isothermal expansion of an ideal gas:
\[
W = nRT\ln\left(\frac{V_2}{V_1}\right)
\]
Substituting values:
\[
W = 2 \times 8.314 \times 300 \times \ln\left(\frac{20}{10}\right)
\]
\[
W = 2 \times 8.314 \times 300 \times 0.693 = 3457.2\,\text{J} \approx 3.46\,\text{kJ}
\]
So the correct option is B.