Q. Ten moles of an ideal gas at constant temperature 600 K is compressed from 100 L to 10 L. The work done in the process is
A) $4.11 \times 10^{4}$ J
B) $- 4.11 \times 10^{4}$J
C) $11.4 \times 10^{4}$J
D) $- 11.4 \times 10^{4}$J
Explanation
(d) $W = 2.3026\ nRT\log_{10}\left( \frac{V_{2}}{V_{1}} \right)$
$= 2.3026 \times 10 \times 9.3 \times 600\log_{10}\left( \frac{10}{100} \right)$
$= - 11.4 \times 10^{4}\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.