Two thermally insulated vessels 1 and 2 are f | vedclass

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(C) The total number of moles $n$ remains conserved when the valve is opened.$n = n_1 + n_2$Using the ideal gas law $PV = nRT$,we have $n = \frac{PV}{

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$\frac{T_1 T_2 (P_1 V_1 + P_2 V_2)}{P_1 V_1 T_2 + P_2 V_2 T_1}$

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(C) The total number of moles $n$ remains conserved when the valve is opened.$n = n_1 + n_2$Using the ideal gas law $PV = nRT$,we have $n = \frac{PV}{RT}$.Thus,$\frac{P(V_1 + V_2)}{RT} = \frac{P_1 V_1}{RT_1} + \frac{P_2 V_2}{RT_2}$.Since the vessels are thermally insulated,the total internal energy is conserved. For an ideal gas,$U = \frac{f}{2} nRT$. Since the gas is the same (air),$f$ is constant.Total internal energy $U_i = U_f \Rightarrow n_1 T_1 + n_2 T_2 = n T$.Substituting $n_1 = \frac{P_1 V_1}{RT_1}$ and $n_2 = \frac{P_2 V_2}{RT_2}$,we get $\frac{P_1 V_1}{R} + \frac{P_2 V_2}{R} = nT$.From the mole conservation,$n = \frac{P_1 V_1}{RT_1} + \frac{P_2 V_2}{RT_2} = \frac{P_1 V_1 T_2 + P_2 V_2 T_1}{RT_1 T_2}$.Equating the energy expressions: $\frac{P_1 V_1 + P_2 V_2}{R} = \left( \frac{P_1 V_1 T_2 + P_2 V_2 T_1}{RT_1 T_2} \right) T$.Solving for $T$: $T = \frac{T_1 T_2 (P_1 V_1 + P_2 V_2)}{P_1 V_1 T_2 + P_2 V_2 T_1}$.

Two thermally insulated vessels $1$ and $2$ are filled with air at temperatures $T_1, T_2$,pressures $P_1, P_2$,and volumes $V_1, V_2$ respectively. If the valve connecting them is opened,what is the temperature of the equilibrium mixture?

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