Three moles of a gas at a temperature T is he | vedclass

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(B) Number of moles,$n = 3$. Initial volume,$V_i = V$. Final volume,$V_f = 3V$. Ratio of specific heats,$\gamma = \frac{C_p}{C_v}$. Change in internal

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$\frac{6 RT}{\gamma-1}$

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(B) Number of moles,$n = 3$. Initial volume,$V_i = V$. Final volume,$V_f = 3V$. Ratio of specific heats,$\gamma = \frac{C_p}{C_v}$. Change in internal energy is given by $\Delta U = n C_v \Delta T$. We know that $C_v = \frac{R}{\gamma - 1}$. At constant pressure,according to Charles's Law,$\frac{V_1}{T_1} = \frac{V_2}{T_2}$. Substituting the values,$\frac{V}{T} = \frac{3V}{T_2}$,which gives $T_2 = 3T$. Change in temperature,$\Delta T = T_2 - T_1 = 3T - T = 2T$. Now,substituting these into the internal energy formula: $\Delta U = n \left( \frac{R}{\gamma - 1} \right) \Delta T = 3 \left( \frac{R}{\gamma - 1} \right) (2T) = \frac{6RT}{\gamma - 1}$.

Three moles of a gas at a temperature $T$ is heated to thrice its volume by keeping the pressure constant. If $\gamma$ is the ratio of specific heats,then the increase in internal energy of the gas is

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