The volume of a gas is reduced adiabatically | vedclass

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(A) For an adiabatic process, the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = 	ext{constant}$.Let the initial vo

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$(300) 	imes 4^{0.4} \, K$

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(A) For an adiabatic process, the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = 	ext{constant}$.Let the initial volume be $V_0$ and the initial temperature be $T_0 = 27 + 273 = 300 \, K$.The final volume is $V = V_0 / 4$.Given $\gamma = 1.4$, then $\gamma - 1 = 0.4$.Using the adiabatic relation: $T_0 V_0^{\gamma-1} = T V^{\gamma-1}$.$300 	imes V_0^{0.4} = T 	imes (V_0 / 4)^{0.4}$.$300 	imes V_0^{0.4} = T 	imes V_0^{0.4} 	imes (1/4)^{0.4}$.$300 = T 	imes (4^{-1})^{0.4}$.$300 = T 	imes 4^{-0.4}$.$T = 300 	imes 4^{0.4} \, K$.

The volume of a gas is reduced adiabatically to $(1/4)^{th}$ of its initial volume. If the initial temperature is $27\,^{\circ}C$ and $\gamma = 1.4$, the new temperature is:

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