Two identical adiabatic vessels are filled wi | vedclass

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(D) For an ideal gas,the internal energy $U$ is given by $U = \frac{f}{2} nRT = \frac{f}{2} PV$. Since the vessels are identical,$V_1 = V_2 = V$. Thus

Vedclass · Accepted Answer

$\frac{U_{01}}{U_{02}} = \frac{P_1}{P_2}, U_{f1} = U_{f2}$

Vedclass · Answer

(D) For an ideal gas,the internal energy $U$ is given by $U = \frac{f}{2} nRT = \frac{f}{2} PV$. Since the vessels are identical,$V_1 = V_2 = V$. Thus,$\frac{U_{01}}{U_{02}} = \frac{P_1 V}{P_2 V} = \frac{P_1}{P_2}$.When the vessels are connected,gas flows from the high-pressure vessel to the low-pressure vessel until the pressures equalize. At equilibrium,the final pressure $P_f$ is the same in both vessels. Since the vessels are identical and the pipe is non-conducting,the final state will have equal pressure and equal temperature in both vessels. Therefore,$U_{f1} = \frac{f}{2} P_f V$ and $U_{f2} = \frac{f}{2} P_f V$,which implies $U_{f1} = U_{f2}$.

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