Q. 5 mole of an ideal gas with $(\gamma = 7/5)$ initially at STP are compressed adiabatically so that its temperature becomes $400℃$. The increase in the internal energy of gas in kJ is
A) 21.55
B) 41.55
C) 65.55
D) 50.55
Explanation
(b) Here, $n = 5,\ \gamma = \frac{7}{5},\ T_{1} = 0℃,\ T_{2} = 400℃$
$dU = \frac{nRdT}{\frac{7}{5} - 1}$
$dU = \frac{5 \times 8.31 \times (400 - 0)}{\frac{7}{5} - 1} = 41550\ J$
$dU = 41.55\ kJ$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.