Q.An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume $V_1$ and contains ideal gas at pressure $P_1$ and temperature $T_1$. The other chamber has volume $V_2$ and contains ideal gas at pressure $P_2$ and temperature $T_2$. If the partition is removed without doing any work on the gases, calculate the final equilibrium temperature of the container.

Step 1. Since the container is insulated (no heat lost to outside) and the partition is removed without doing any external work, the total internal energy of the combined gas is conserved -- it simply equals the sum of the two chambers' internal energies before mixing.

Step 2. The number of moles in each chamber (from the ideal gas law) is $n_1=\dfrac{P_1V_1}{RT_1}$ and $n_2=\dfrac{P_2V_2}{RT_2}$. Assuming both chambers hold the same gas (so the same molar specific heat capacity $C_v$ applies throughout), conservation of internal energy gives $n_1C_vT_1+n_2C_vT_2=(n_1+n_2)C_vT_f,$ so $T_f=\dfrac{n_1T_1+n_2T_2}{n_1+n_2}.$

Step 3. Now $n_1T_1=\dfrac{P_1V_1}{RT_1}\times T_1=\dfrac{P_1V_1}{R}$, and similarly $n_2T_2=\dfrac{P_2V_2}{R}$; so the numerator is $\dfrac{P_1V_1+P_2V_2}{R}$.

Step 4. And $n_1+n_2=\dfrac{P_1V_1}{RT_1}+\dfrac{P_2V_2}{RT_2}=\dfrac{P_1V_1T_2+P_2V_2T_1}{RT_1T_2}$. Dividing the Step 3 numerator by this denominator (the $R$'s cancel): $T_f=\dfrac{(P_1V_1+P_2V_2)/R}{(P_1V_1T_2+P_2V_2T_1)/(RT_1T_2)}=\dfrac{T_1T_2(P_1V_1+P_2V_2)}{P_1V_1T_2+P_2V_2T_1}.$

Use conservation of internal energy ($n_1C_vT_1+n_2C_vT_2=(n_1+n_2)C_vT_f$) and substitute $n_i=P_iV_i/(RT_i)$ throughout.

• Trying to apply the calorimetry mixing formula ($T_f=(m_1s_1T_1+m_2s_2T_2)/(m_1s_1+m_2s_2)$) directly -- that formula is for two DIFFERENT masses/substances at a common final state; here it must be derived from scratch via mole counts and the ideal gas law, since the chambers are described by P, V, T rather than mass and specific heat.