What is the final temperature of 0.10 , mol m | vedclass

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(A) For an adiabatic process,the first law of thermodynamics is given by $\Delta U = q + W$. Since $q = 0$,$\Delta U = W$. Given that the gas performs

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$250$

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(A) For an adiabatic process,the first law of thermodynamics is given by $\Delta U = q + W$. Since $q = 0$,$\Delta U = W$. Given that the gas performs work,$W = -75 \, cal$. The change in internal energy for a monoatomic ideal gas is $\Delta U = n C_v \Delta T$. For a monoatomic gas,$C_v = \frac{3}{2} R$. Initial temperature $T_1 = 227 + 273 = 500 \, K$. Substituting the values: $-75 = 0.10 	imes (\frac{3}{2} 	imes 2) 	imes (T_2 - 500)$. $-75 = 0.10 	imes 3 	imes (T_2 - 500)$. $-75 = 0.3 	imes (T_2 - 500)$. $-250 = T_2 - 500$. $T_2 = 500 - 250 = 250 \, K$.

What is the final temperature of $0.10 \, mol$ monoatomic ideal gas that performs $75 \, cal$ of work adiabatically if the initial temperature is $227 \, ^oC$? .......$K$ (use $R = 2 \, cal \cdot K^{-1} \cdot mol^{-1}$)

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