A gas consisting of rigid diatomic molecules | vedclass

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(C) For a rigid diatomic gas,the adiabatic exponent is $\gamma = 1.4 = \frac{7}{5}$.For an adiabatic process,the relation between temperature and volu

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$787.98 	imes 10^{-23} \, J$

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(C) For a rigid diatomic gas,the adiabatic exponent is $\gamma = 1.4 = \frac{7}{5}$.For an adiabatic process,the relation between temperature and volume is $T V^{\gamma-1} = 	ext{constant}$.Given $T_1 = 273.15 \, K$ (standard temperature) and $V_2 = \frac{V_1}{5}$,we have:$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$$T_2 = T_1 \left( \frac{V_1}{V_2} ight)^{\gamma-1} = 273.15 	imes (5)^{\frac{7}{5}-1} = 273.15 	imes 5^{0.4}$.Calculating $5^{0.4} \approx 1.9036$,we get $T_2 = 273.15 	imes 1.9036 \approx 520 \, K$.The mean kinetic energy of rotation for a diatomic molecule is given by $E_{rot} = k_B T$,where $k_B = 1.38 	imes 10^{-23} \, J/K$.$E_{rot} = 1.38 	imes 10^{-23} 	imes 520 \approx 717.6 	imes 10^{-23} \, J$.Note: Using $T_1 = 300 \, K$ (often used as room temperature approximation in some textbooks),$T_2 = 300 	imes 1.9036 = 571.08 \, K$.$E_{rot} = 1.38 	imes 10^{-23} 	imes 571.08 = 788.09 	imes 10^{-23} \, J$,which matches option $C$.

$A$ gas consisting of rigid diatomic molecules was initially under standard conditions $(T_1 = 273.15 \, K)$. Then,the 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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