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(D) Given: $\gamma = 1.5$,$P_1 = 200 \, kPa = 2 	imes 10^5 \, Pa$,$V_1 = 1200 \, cm^3 = 1200 	imes 10^{-6} \, m^3 = 1.2 	imes 10^{-3} \, m^3$,$V_2

Vedclass · Accepted Answer

$480$

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(D) Given: $\gamma = 1.5$,$P_1 = 200 \, kPa = 2 	imes 10^5 \, Pa$,$V_1 = 1200 \, cm^3 = 1200 	imes 10^{-6} \, m^3 = 1.2 	imes 10^{-3} \, m^3$,$V_2 = 300 \, cm^3 = 300 	imes 10^{-6} \, m^3 = 0.3 	imes 10^{-3} \, m^3$.For an adiabatic process,$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$.$P_2 = P_1 (V_1 / V_2)^{\gamma} = 200 	imes (1200 / 300)^{1.5} = 200 	imes (4)^{1.5} = 200 	imes 8 = 1600 \, kPa = 16 	imes 10^5 \, Pa$.The work done by the gas in an adiabatic process is $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$.$W = \frac{(2 	imes 10^5 	imes 1.2 	imes 10^{-3}) - (16 	imes 10^5 	imes 0.3 	imes 10^{-3})}{1.5 - 1} = \frac{240 - 480}{0.5} = \frac{-240}{0.5} = -480 \, J$.The absolute value of the work done is $|W| = 480 \, J$.

$A$ sample of gas with $\gamma=1.5$ is taken through an adiabatic process in which the volume is compressed from $1200 \, cm^3$ to $300 \, cm^3$. If the initial pressure is $200 \, kPa$,find the absolute value of the work done by the gas in the process in Joules.

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