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(B) For an adiabatic process,the work done $W$ on the gas is given by $W = \frac{nR\Delta T}{1 - \gamma}$.Here,$n = 1 	ext{ kilo mole} = 1000 	ext{

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diatomic

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(B) For an adiabatic process,the work done $W$ on the gas is given by $W = \frac{nR\Delta T}{1 - \gamma}$.Here,$n = 1 	ext{ kilo mole} = 1000 	ext{ moles}$,$R = 8.3 \ J \ mol^{-1} K^{-1}$,$\Delta T = 7 \ K$,and $W = -146 \ kJ = -146000 \ J$ (work is done on the gas).Substituting the values: $-146000 = \frac{1000 	imes 8.3 	imes 7}{1 - \gamma}$.$1 - \gamma = -\frac{58100}{146000} \approx -0.3979 \approx -0.4$.$1 - \gamma = -0.4 \Rightarrow \gamma = 1.4$.Since $\gamma = 1.4$ for a diatomic gas,the gas is diatomic.

The work of $146 \ kJ$ is performed in order to compress one kilo mole of gas adiabatically,and in this process,the temperature of the gas increases by $7 ^\circ C$. The gas is $(R = 8.3 \ J \ mol^{-1} K^{-1})$.

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