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Rigorous reason behind Internalinternal energy change being zero while mixing

A container is divided in two parts: One part contains Oxygen Gas ($n_{1}$ moles,at Temperature $T1$) and the other part contains helium gas ($n_{2}$ moles, at temperature $T2$). The partition separating them is removed. We need to find the final temperature of the mixture $(T)$.

A container is divided in two parts: one part contains oxygen gas $(n_1$ moles, at temperature $T_1)$ and the other part contains helium gas $(n_2$ moles, at temperature $T_2).$ The partition separating them is removed. Find the final temperature of the mixture $(T).$

While solving this (multiple choice exam) question, since I was constrained by time, I started to look for hints within the question.I I realized the question has explicitly mentioned the gases:, i.e. it gave us information about their degrees of freedom.  (since oxygen is diatomic and helium is monoatomic). So my instinct was that this had to do something with $Cv$.$C_V.$ At the time, I hypothesized $\Delta{U}_{sys}=0$.$\Delta{U}_\mathrm{sys} = 0.$ Thus: $$n_{1}*(5R/2)(T-T1) + n_{2}*(3R/2)(T-T2)=0$$

$$n_1\frac{5R}{2}(T - T_1) + n_2\frac{3R}{2}(T - T_2) = 0$$ $$T=\dfrac{5n_{1}T1+3n_2T2}{5n1+3n2}$$$$T = \frac{5n_1T_1 + 3n_2T_2}{5n_1 + 3n_2}$$

Which was indeed one of the options, and infactin fact, the correct one.

However, Isis there any rigorous method/reasoning that tells us why exactlyexactly is it that $\Delta{U}_{sys}=0?$$\Delta{U}_\mathrm{sys} = 0?$ I cantcan't think of a satisfactory explanation.I I tried to analyze it using the first law of thermodynamics, but that clearly seems to be the wrong approach.

Rigorous reason behind Internal energy change being zero while mixing

A container is divided in two parts: One part contains Oxygen Gas ($n_{1}$ moles,at Temperature $T1$) and the other part contains helium gas ($n_{2}$ moles, at temperature $T2$). The partition separating them is removed. We need to find the final temperature of the mixture $(T)$.

While solving this (multiple choice exam) question, since I was constrained by time, I started to look for hints within the question.I realized the question has explicitly mentioned the gases: i.e it gave us information about their degrees of freedom.(since oxygen is diatomic and helium is monoatomic). So my instinct was that this had to do something with $Cv$. At the time, I hypothesized $\Delta{U}_{sys}=0$. Thus: $$n_{1}*(5R/2)(T-T1) + n_{2}*(3R/2)(T-T2)=0$$ $$T=\dfrac{5n_{1}T1+3n_2T2}{5n1+3n2}$$

Which was indeed one of the options, and infact, the correct one.

However, Is there any rigorous method/reasoning that tells us why exactly is it that $\Delta{U}_{sys}=0?$ I cant think of a satisfactory explanation.I tried to analyze it using the first law of thermodynamics, but that clearly seems to be the wrong approach.

Rigorous reason behind internal energy change being zero while mixing

A container is divided in two parts: one part contains oxygen gas $(n_1$ moles, at temperature $T_1)$ and the other part contains helium gas $(n_2$ moles, at temperature $T_2).$ The partition separating them is removed. Find the final temperature of the mixture $(T).$

While solving this (multiple choice exam) question, since I was constrained by time, I started to look for hints within the question. I realized the question has explicitly mentioned the gases, i.e. it gave us information about their degrees of freedom  (since oxygen is diatomic and helium is monoatomic). So my instinct was that this had to do something with $C_V.$ At the time, I hypothesized $\Delta{U}_\mathrm{sys} = 0.$ Thus:

$$n_1\frac{5R}{2}(T - T_1) + n_2\frac{3R}{2}(T - T_2) = 0$$ $$T = \frac{5n_1T_1 + 3n_2T_2}{5n_1 + 3n_2}$$

Which was indeed one of the options, and in fact, the correct one.

However, is there any rigorous method/reasoning that tells us why exactly is it that $\Delta{U}_\mathrm{sys} = 0?$ I can't think of a satisfactory explanation. I tried to analyze it using the first law of thermodynamics, but that clearly seems to be the wrong approach.

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Rigorous reason behind Internal energy change being zero while mixing

A container is divided in two parts: One part contains Oxygen Gas ($n_{1}$ moles,at Temperature $T1$) and the other part contains helium gas ($n_{2}$ moles, at temperature $T2$). The partition separating them is removed. We need to find the final temperature of the mixture $(T)$.

While solving this (multiple choice exam) question, since I was constrained by time, I started to look for hints within the question.I realized the question has explicitly mentioned the gases: i.e it gave us information about their degrees of freedom.(since oxygen is diatomic and helium is monoatomic). So my instinct was that this had to do something with $Cv$. At the time, I hypothesized $\Delta{U}_{sys}=0$. Thus: $$n_{1}*(5R/2)(T-T1) + n_{2}*(3R/2)(T-T2)=0$$ $$T=\dfrac{5n_{1}T1+3n_2T2}{5n1+3n2}$$

Which was indeed one of the options, and infact, the correct one.

However, Is there any rigorous method/reasoning that tells us why exactly is it that $\Delta{U}_{sys}=0?$ I cant think of a satisfactory explanation.I tried to analyze it using the first law of thermodynamics, but that clearly seems to be the wrong approach.