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(B) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$.Given: $n = 2 \ mol$,$T_1 =

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$(a) 189 \ K, (b) -2.7 \ kJ$

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(B) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$.Given: $n = 2 \ mol$,$T_1 = 27^{\circ}C = 300 \ K$,$V_1 = V$,$V_2 = 2V$.For a monoatomic gas,the adiabatic index $\gamma = 5/3$,so $\gamma - 1 = 2/3$.$(a)$ Substituting the values: $300 	imes V^{2/3} = T_2 	imes (2V)^{2/3}$.$T_2 = 300 / (2^{2/3}) \approx 300 / 1.5874 \approx 189 \ K$.$(b)$ The change in internal energy is given by $\Delta U = n C_v \Delta T$.For a monoatomic gas,$C_v = (3/2)R$.$\Delta U = 2 	imes (3/2) 	imes 8.314 	imes (189 - 300) \approx 3 	imes 8.314 	imes (-111) \approx -2768 \ J \approx -2.7 \ kJ$.

Two moles of an ideal monoatomic gas occupy a volume $V$ at $27^{\circ}C$. The gas expands adiabatically to a volume $2V$. Calculate $(a)$ the final temperature of the gas and $(b)$ the change in its internal energy.

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