For an adiabatic compression of 1 text{ kmol} | vedclass

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(B) According to the first law of thermodynamics,$\Delta Q = \Delta E_{int} + \Delta W$.For an adiabatic process,$\Delta Q = 0$.Therefore,$0 = \Delta

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(B) According to the first law of thermodynamics,$\Delta Q = \Delta E_{int} + \Delta W$.For an adiabatic process,$\Delta Q = 0$.Therefore,$0 = \Delta E_{int} + \Delta W$,which implies $\Delta E_{int} = -\Delta W$.The change in internal energy is given by $\Delta E_{int} = n C_V \Delta T$,where $n$ is the number of moles.Given $n = 1 	ext{ kmol} = 1000 	ext{ mol}$,$\Delta W = -146 	ext{ kJ} = -146 	imes 10^3 	ext{ J}$ (work done on the gas is negative in the system's perspective for expansion,but here compression is done,so $\Delta W = -146 	ext{ kJ}$),and $\Delta T = 7 	ext{ K}$.$C_V = \frac{-\Delta W}{n \Delta T} = \frac{-(-146 	imes 10^3)}{1000 	imes 7} = \frac{146}{7} \approx 20.86 	ext{ J mol}^{-1} 	ext{ K}^{-1}$.For a diatomic gas,$C_V = \frac{5}{2} R = 2.5 	imes 8.314 \approx 20.78 	ext{ J mol}^{-1} 	ext{ K}^{-1}$.Since the calculated $C_V$ matches the value for a diatomic gas,the gas is diatomic.

For an adiabatic compression of $1 \text{ kmol}$ of gas,$146 \text{ kJ}$ of work is done. During this process,the temperature of the gas increases by $7 \text{ °C}$. The gas is ........

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