Q. A perfect gas goes from state $A$ to state $B$ by absorbing $8 \times 10^{5}\ J$ of heat and doing $6.5 \times 10^{5}\ J$ of external work. It is now transferred between the same two states in another process in which it absorbs $10^{5}$ J of heat. In the second process,

A) Work done on gas is $10^{5}$ J
B) Work done on gas is $0.5 \times 10^{5}\ J$
C) Work done by gas is $10^{5}$ J
D) Work done by gas $0.5 \times 10^{5}\ J$
Explanation

(b) $dU = dQ - dW = 8 \times 10^{5} - 6.5 \times 10^{5} = 1.5 \times 10^{5}\ J$
In the 2nd process, $dU$ remains the same
$\therefore\ dW = dQ - dU = 10^{5} - 1.5 \times 10^{5} = - 0.5 \times 10^{5}\ J$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

Want more practice?
Practice Thermodynamics MCQs