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(A) For an adiabatic process,the work done on the gas is given by $W = \frac{nR \Delta T}{\gamma - 1}$.Given: $W = 146 \, kJ = 146 	imes 10^3 \, J$,$

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diatomic

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(A) For an adiabatic process,the work done on the gas is given by $W = \frac{nR \Delta T}{\gamma - 1}$.Given: $W = 146 \, kJ = 146 	imes 10^3 \, J$,$n = 1 \, kmol = 10^3 \, mol$,$\Delta T = 7 \, K$,and $R = 8.3 \, J \, mol^{-1} \, K^{-1}$.Substituting the values into the formula:$146 	imes 10^3 = \frac{10^3 	imes 8.3 	imes 7}{\gamma - 1}$$146 = \frac{58.1}{\gamma - 1}$$\gamma - 1 = \frac{58.1}{146} \approx 0.3979 \approx 0.4$$\gamma = 1 + 0.4 = 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 a gas adiabatically,and in this process,the temperature of the gas increases by $7 \, ^oC$. The gas is $(R = 8.3 \, J \, mol^{-1} \, K^{-1})$.

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