One gram mole of a diatomic gas (gamma = 1.4) | vedclass

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Question

(A) For an adiabatic process,the work done $W$ is given by the formula: $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.Given: $n = 1 \, 	ext{mole}$,$R = 8.31

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$2077.5 \, J$

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(A) For an adiabatic process,the work done $W$ is given by the formula: $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.Given: $n = 1 \, 	ext{mole}$,$R = 8.31 \, J/(mol \cdot K)$,$\gamma = 1.4$,$T_1 = 27^{\circ}C = 300 \, K$,and $T_2 = 127^{\circ}C = 400 \, K$.Substituting the values: $W = \frac{1 	imes 8.31 	imes (300 - 400)}{1.4 - 1}$.$W = \frac{8.31 	imes (-100)}{0.4} = \frac{-831}{0.4} = -2077.5 \, J$.The negative sign indicates that work is done on the gas during compression. The magnitude of the work done is $2077.5 \, J$.

One gram mole of a diatomic gas $(\gamma = 1.4)$ is compressed adiabatically so that its temperature rises from $27^{\circ}C$ to $127^{\circ}C$. The work done will be:

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