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(A) For a diatomic gas with rigid molecules,the degrees of freedom $f = 5$ and the adiabatic index $\gamma = 1 + 2/f = 7/5 = 1.4$.Initial temperature

Vedclass · Accepted Answer

$46.87$

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(A) For a diatomic gas with rigid molecules,the degrees of freedom $f = 5$ and the adiabatic index $\gamma = 1 + 2/f = 7/5 = 1.4$.Initial temperature $T_1 = 20 + 273 = 293\,K$.For an adiabatic process,$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Given $V_2 = V_1 / 10$,we have $T_2 = T_1 (V_1 / V_2)^{\gamma-1} = 293 	imes (10)^{1.4-1} = 293 	imes 10^{0.4}$.Using $10^{0.4} \approx 2.5118$,$T_2 = 293 	imes 2.5118 \approx 735.96\,K$.The change in internal energy is $\Delta U = n C_v \Delta T = n (fR/2) (T_2 - T_1)$.Using $n = 5$,$f = 5$,and $R = 8.314\,J/(mol\cdot K)$:$\Delta U = 5 	imes (5 	imes 8.314 / 2) 	imes (735.96 - 293) = 12.5 	imes 8.314 	imes 442.96 \approx 46056\,J = 46.056\,kJ$.Rounding to the nearest integer,$X \approx 46$.

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