A diatomic gas of volume 2 m^3 at pressure 2 | vedclass

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(C) For an adiabatic process,the work done is given by the formula: $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$.For a diatomic gas,the adiabatic index

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$-7.4 	imes 10^5 \ J$

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(C) For an adiabatic process,the work done is given by the formula: $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$.For a diatomic gas,the adiabatic index $\gamma = 1.4$.Given: $P_1 = 2 	imes 10^5 \ N \ m^{-2}$,$V_1 = 2 \ m^3$,$V_2 = 0.5 \ m^3$.Using the adiabatic relation $P_1 V_1^\gamma = P_2 V_2^\gamma$,we find $P_2 = P_1 \left(\frac{V_1}{V_2}ight)^\gamma = 2 	imes 10^5 	imes \left(\frac{2}{0.5}ight)^{1.4} = 2 	imes 10^5 	imes (4)^{1.4}$.Given $(4)^{1.4} = 6.96$,so $P_2 = 2 	imes 10^5 	imes 6.96 = 13.92 	imes 10^5 \ N \ m^{-2}$.Now,substitute the values into the work formula:$W = \frac{(2 	imes 10^5 	imes 2) - (13.92 	imes 10^5 	imes 0.5)}{1.4 - 1}$$W = \frac{4 	imes 10^5 - 6.96 	imes 10^5}{0.4} = \frac{-2.96 	imes 10^5}{0.4} = -7.4 	imes 10^5 \ J$.The negative sign indicates that work is done on the gas during compression.

$A$ diatomic gas of volume $2 \ m^3$ at pressure $2 \times 10^5 \ N \ m^{-2}$ is compressed adiabatically to a volume $0.5 \ m^3$. The work done in this process is,$[$Use $4^{1.4} = 6.96]$

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