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(D) Given,initial temperature of $1$ mole of $N_2$ gas,$T_1 = 300 \,K$. Initial pressure of gas,$p_1 = p$. Final pressure of gas,$p_2 = 10p$. For a ri

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$570$

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(D) Given,initial temperature of $1$ mole of $N_2$ gas,$T_1 = 300 \,K$. Initial pressure of gas,$p_1 = p$. Final pressure of gas,$p_2 = 10p$. For a rigid diatomic gas,the adiabatic index $\gamma = 7/5 = 1.4$. Using the adiabatic relation between pressure and temperature: $T_1^\gamma p_1^{1-\gamma} = T_2^\gamma p_2^{1-\gamma}$. Rearranging for $T_2$: $(T_2/T_1)^\gamma = (p_1/p_2)^{1-\gamma} = (p_2/p_1)^{\gamma-1}$. $T_2 = T_1 	imes (p_2/p_1)^{(\gamma-1)/\gamma} = 300 	imes (10p/p)^{(1.4-1)/1.4} = 300 	imes 10^{2/7}$. Since $10^{2/7} = (10^2)^{1/7} = 100^{1/7} = 1.9$,we have $T_2 = 300 	imes 1.9 = 570 \,K$.

One mole of nitrogen gas being initially at a temperature of $T_0 = 300 \,K$ is adiabatically compressed to increase its pressure $10$ times. The final gas temperature after compression is (Assume,nitrogen gas molecules as rigid diatomic and $100^{1/7} = 1.9$) (in $\,K$)

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