Two thermally insulated vessels 1 and 2 are f | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The total number of moles $n$ is conserved when the valve is opened.$n = n_1 + n_2$Using the ideal gas law $PV = nRT$,we have $n = \frac{PV}{RT}$.So,$\frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2} = \frac{P(V_1 + V_2)}{R T}$Since the system is thermally insulated,the total internal energy is conserved. For an ideal gas,$U = \frac{f}{2} nRT$. Thus,$n_1 T_1 + n_2 T_2 = (n_1 + n_2) T$.Substituting $n = \frac{PV}{RT}$,we get $\frac{P_1 V_1}{R T_1} T_1 + \frac{P_2 V_2}{R T_2} T_2 = (n_1 + n_2) T$.$n_1 + n_2 = \frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2} = \frac{P_1 V_1 T_2 + P_2 V_2 T_1}{R T_1 T_2}$.Also,the final pressure $P$ is given by $P = \frac{(n_1 + n_2)RT}{V_1 + V_2}$.Substituting the values into the energy conservation equation: $P_1 V_1 + P_2 V_2 = (n_1 + n_2) RT$.Therefore,$T = \frac{P_1 V_1 + P_2 V_2}{n_1 + n_2} = \frac{(P_1 V_1 + P_2 V_2) T_1 T_2}{P_1 V_1 T_2 + P_2 V_2 T_1}$.

Two thermally insulated vessels $1$ and $2$ are filled with air at temperatures $(T_1, T_2)$,volumes $(V_1, V_2)$,and pressures $(P_1, P_2)$ respectively. If the valve joining the two vessels is opened,the temperature inside the vessel at equilibrium will be

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