Q 11-11-032JEE MainMedium
Find the work done, in joules, when $2$ mol of an ideal gas expand isothermally at $300$ K from $10$ L to $20$ L. ($R = 8.3$ J/mol K, $\ln 2 = 0.693$; give the answer to the nearest whole number)
Numerical value type. Enter your answer.
Answer: 3451
Concept. Isothermal work: $W = nRT\ln\dfrac{V_2}{V_1}$.
Step 1. Substitute. $2 \times 8.3 \times 300 \times 0.693 \approx 3451$ J.
Answer. $3451$.
Solution by Sreeraj P, M.Sc Physics