Q. At N.T.P. one mole of diatomic gas is compressed adiabatically to half of its volume, $\gamma = 1.41$. The work done on gas will be
A) $1280\ J$
B) $1610\ J$
C) $1815\ J$
D) $2025\ J$
Explanation
(c) $T_{2} = T_{1}\left( \frac{V_{1}}{V_{2}} \right)^{\gamma - 1} = 273(2)^{0.41} = 273 \times 1.328 = 363\ K$
$W = \frac{R(T_{1} - T_{2})}{\gamma - 1} = \frac{8.31(273 - 363)}{1.41 - 1} = - 1824$
$\Rightarrow |W| \approx 1815\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.