A monatomic gas at 630 ,K expands adiabatical | vedclass

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(D) For a monatomic gas, the degrees of freedom $f=3$. The adiabatic index $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$. For an adiabati

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$70$

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(D) For a monatomic gas, the degrees of freedom $f=3$. The adiabatic index $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$. For an adiabatic process, the relationship between temperature $T$ and volume $V$ is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$. Given $T_1 = 630 \,K$, $V_1 = V$, and $V_2 = 27V$. Substituting these values: $630 	imes V^{\frac{5}{3}-1} = T_2 	imes (27V)^{\frac{5}{3}-1}$ $630 = T_2 	imes (27)^{\frac{2}{3}}$ $630 = T_2 	imes (3^3)^{\frac{2}{3}}$ $630 = T_2 	imes 3^2$ $630 = T_2 	imes 9$ $T_2 = \frac{630}{9} = 70 \,K$.

$A$ monatomic gas at $630 \,K$ expands adiabatically to $27$ times its initial volume. The final temperature of the gas is (in $\,K$)

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