Question

A gas consisting of rigid diatomic molecules was initially under standard condition, then gas was compressed adiabatically to one-fifth of its initial volume. What will be the mean kinetic energy of a rotating molecule in the final state?
$[$ Take, $\gamma$ for diatomic gas $=\frac{7}{5}]$

• A. $1.44J$
• B. $4.55J$
• C. $787.98\times 10^{-23}J$
• D. $757.3\times 10^{-23}J$

Answer 1

Answer: (C)

Given, for diatomic gas, $\gamma =\frac{7}{5}$
For adiabatic process, $T{V}^{\gamma -1}=$ constant
$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$
Given $V_2=\frac{V_1}{5}\text{ and }{T}_1=27^{\circ }C=300K$
(at standard conditions)
$(300)V_1^{\frac{7}{5}-1}=T_2{\left(\frac{V_1}{5}\right)}^{\frac{7}{5}-1}$
$\Rightarrow T_2=\frac{300\times V_1^{2/5}}{V_1^{\frac{2}{5}}\times {\left(\frac{1}{5}\right)}^{2/5}}=\frac{300}{5^{\frac{-2}{5}}}=300\times 5^{2/5}$
$=300\times 1.903=571K$
Mean kinetic energy of rotating molecules
$=kT=1.38\times 10^{-23}\times 571=787.98\times 10^{-23}J$