1 , text{kmol} of an ideal gas is compressed | vedclass

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(D) For an adiabatic process,the heat exchange $Q = 0$. According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$. Since $Q = 0$,$

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Diatomic

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(D) For an adiabatic process,the heat exchange $Q = 0$. According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$. Since $Q = 0$,$\Delta U = -\Delta W$.Given that work is done on the gas,$\Delta W = -146 	imes 10^3 \, 	ext{J}$,so $\Delta U = 146 	imes 10^3 \, 	ext{J}$.The change in internal energy is given by $\Delta U = n C_v \Delta T$,where $n = 10^3 \, 	ext{mol}$ and $\Delta T = 7 \, 	ext{K}$.$C_v = \frac{R}{\gamma - 1}$.Substituting the values: $146 	imes 10^3 = 10^3 	imes \frac{8.3}{\gamma - 1} 	imes 7$.$146 = \frac{58.1}{\gamma - 1} \implies \gamma - 1 = \frac{58.1}{146} \approx 0.3979$.$\gamma \approx 1.4$.Since $\gamma = 1.4$ for a diatomic gas,the gas is diatomic.

$1 \, \text{kmol}$ of an ideal gas is compressed adiabatically,requiring $146 \, \text{kJ}$ of work. During this process,the temperature of the gas increases by $7^o \text{C}$. The gas is: $(R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1})$

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