Question 10: Thermodynamics & KTG

Question

A gas consisting of a 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?

Accepted Answer

For diatomic gas, $$\gamma=1.4=\frac{7}{5}$$
For adiabatic process, $$T V^{\gamma-1}=$$ constant
$$\begin{aligned}
T_1 V_1^{\gamma-1} & =T_2 V_2^{\gamma-1} \
(300) V_1^{\frac{7}{5}-1} & =T_2\left(\frac{V_1}{5}ight)^{\frac{7}{5}-1} \
\Rightarrow \quad T_2 & =\frac{300 	imes V_1^{2 / 5}}{V_1^{\frac{2}{5}} 	imes\left(\frac{1}{5}ight)^{2 / 5}}=\frac{300}{5^{\frac{-2}{5}}}=300 	imes 5^{2 / 5} \
& =300 	imes 1.903=571
\end{aligned}$$
Mean kinetic energy of rotating molecules
$$=k T=1.38 	imes 10^{-23} 	imes 571=787.98 	imes 10^{-23} \mathrm{~J}$$

Answer

1.44 J

Answer

4.55 J

Answer

$$787.98 	imes 10^{-23} \mathrm{~J}$$

Answer

$$757.3 	imes 10^{-23} \mathrm{~J}$$

Thermodynamics & KTG

A gas consisting of a 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?

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