Two ideal polyatomic gases at temperatures T_ | vedclass

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(B) Let the final temperature of the mixture be $T$. Since there is no loss of energy,the total internal energy of the system is conserved.The interna

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$\frac{n_{1}F_{1}T_{1} + n_{2}F_{2}T_{2}}{n_{1}F_{1} + n_{2}F_{2}}$

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(B) Let the final temperature of the mixture be $T$. Since there is no loss of energy,the total internal energy of the system is conserved.The internal energy of a gas is given by $U = \frac{F}{2} n R T$,where $n$ is the number of moles. Since $n$ is proportional to the number of molecules $N$,we can write $U = \frac{F}{2} N k_{B} T$.For the mixture,the total energy before mixing equals the total energy after mixing:$U_{1} + U_{2} = U_{mix}$$\frac{F_{1}}{2} n_{1} k_{B} T_{1} + \frac{F_{2}}{2} n_{2} k_{B} T_{2} = \frac{F_{1}}{2} n_{1} k_{B} T + \frac{F_{2}}{2} n_{2} k_{B} T$Canceling the common terms $\frac{1}{2} k_{B}$ from both sides:$F_{1} n_{1} T_{1} + F_{2} n_{2} T_{2} = T (F_{1} n_{1} + F_{2} n_{2})$Solving for $T$:$T = \frac{F_{1} n_{1} T_{1} + F_{2} n_{2} T_{2}}{F_{1} n_{1} + F_{2} n_{2}}$

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