Q. When 500 J heat is given to the gas $X$ in an isobaric process its work done comes out as 142.8 J. The gas $X$ is
A) $O_{2}$
B) $NH_{3}$
C) $He$
D) $SO_{2}$
Explanation
(a)
$\because\ \mathrm{\Delta}T = \frac{W}{nR}\therefore Q = nC_{p}(\mathrm{\Delta}T) = nC_{p}\frac{W}{nR} = \frac{C_{p}W}{R}$
$C_{p} = \frac{QR}{W} = \frac{500 \times 2}{142.8} = 7$
$C_{p} = 7$indicates that the gas is diatomic. Thus, it should be $O_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.