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

Edvaya · Accepted Answer

The correct answer is diatomic. **Solution:** For an adiabatic process, the heat exchange $Q = 0$. By the first law of thermodynamics, $W_{	ext{on gas}} = \Delta U = n C_v \Delta T$. Given: Work done on the gas, $W = 146 	ext{ kJ} = 146000 	ext{ J}$ Number of moles, $n = 1 	ext{ kilo mole} = 1000 	ext{ moles}$ Change in temperature, $\Delta T = 7^\circ	ext{C} = 7 	ext{ K}$ Substitute the values into the equation: $146000 = 1000 	imes C_v 	imes 7$ $C_v = \frac{146000}{7000} = \frac{146}{7} 	ext{ J mol}^{-1} 	ext{ K}^{-1}$. We know that $C_v = \frac{R}{\gamma - 1}$. $\frac{R}{\gamma - 1} = \frac{146}{7}$ $\gamma - 1 = \frac{8.3 	imes 7}{146} = \frac{58.1}{146} \approx 0.398 \approx 0.4$ $\gamma \approx 1.4$ A heat capacity ratio $\gamma$ of $1.4$ indicates that the gas is diatomic.

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Physics > Thermodynamics And Kinetic Theory > Thermodynamics

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