One mole of an ideal gas at an initial temper | vedclass

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(C) For an adiabatic process, the first law of thermodynamics states that $Q = \Delta U + W$. Since the process is adiabatic, $Q = 0$, so $\Delta U =

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$(T - 4) \, K$

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(C) For an adiabatic process, the first law of thermodynamics states that $Q = \Delta U + W$. Since the process is adiabatic, $Q = 0$, so $\Delta U = -W$.Given that the work done by the gas is $W = 6 \, R \, 	ext{J}$, we have $\Delta U = -6 \, R$.The change in internal energy for an ideal gas is given by $\Delta U = n C_V \Delta T$.For a gas with adiabatic index $\gamma = \frac{5}{3}$, the molar heat capacity at constant volume is $C_V = \frac{R}{\gamma - 1} = \frac{R}{\frac{5}{3} - 1} = \frac{R}{\frac{2}{3}} = \frac{3}{2} R$.Substituting the values, we get $-6 \, R = 1 	imes \left( \frac{3}{2} R ight) 	imes (T_{final} - T)$.Dividing both sides by $R$, we get $-6 = \frac{3}{2} (T_{final} - T)$.Solving for the temperature change: $T_{final} - T = -6 	imes \frac{2}{3} = -4$.Therefore, $T_{final} = (T - 4) \, K$.

One mole of an ideal gas at an initial temperature of $T \, K$ does $6 \, R \, \text{joules}$ of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $\frac{5}{3}$, the final temperature of the gas will be

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