Two ideal polyatomic gases
Question:

Two ideal polyatomic gases at temperatures $T_{1}$ and $T_{2}$ are mixed so that there is no loss of energy. If $F_{1}$ and $F_{2}, m_{1}$ and $m_{2}, n_{1}$ and $n_{2}$ be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is :

  1. $\frac{\mathrm{n}_{1} \mathrm{~T}_{1}+\mathrm{n}_{2} \mathrm{~T}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$

  2. $\frac{\mathrm{n}_{1} \mathrm{~F}_{1} \mathrm{~T}_{1}+\mathrm{n}_{2} \mathrm{~F}_{2} \mathrm{~T}_{2}}{\mathrm{n}_{1} \mathrm{~F}_{1}+\mathrm{n}_{2} \mathrm{~F}_{2}}$

  3. $\frac{\mathrm{n}_{1} \mathrm{~F}_{1} \mathrm{~T}_{1}+\mathrm{n}_{2} \mathrm{~F}_{2} \mathrm{~T}_{2}}{\mathrm{~F}_{1}+\mathrm{F}_{2}}$

  4. $\frac{\mathrm{n}_{1} \mathrm{~F}_{1} \mathrm{~T}_{1}+\mathrm{n}_{2} \mathrm{~F}_{2} \mathrm{~T}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$


Correct Option: , 2

Solution:

Let the final temperature of the mixture be $\mathrm{T}$.

Since, there is no loss in energy.

$\Delta \mathrm{U}=0$

$\Rightarrow \frac{\mathrm{F}_{1}}{2} \mathrm{n}_{1} \mathrm{R} \Delta \mathrm{T}+\frac{\mathrm{F}_{2}}{2} \mathrm{n}_{2} \mathrm{R} \Delta \mathrm{T}=0$

$\Rightarrow \frac{\mathrm{F}_{1}}{2} \mathrm{n}_{1} \mathrm{R}\left(\mathrm{T}_{1}-\mathrm{T}\right)+\frac{\mathrm{F}_{2}}{2} \mathrm{n}_{2} \mathrm{R}\left(\mathrm{T}_{2}-\mathrm{T}\right)=0$

$\Rightarrow \mathrm{T}=\frac{\mathrm{F}_{1} \mathrm{n}_{1} \mathrm{RT}_{1}+\mathrm{F}_{2} \mathrm{n}_{2} \mathrm{RT} \mathrm{T}_{2}}{\mathrm{~F}_{1} \mathrm{n}_{1} \mathrm{R}+\mathrm{F}_{2} \mathrm{n}_{2} \mathrm{R}} \Rightarrow \frac{\mathrm{F}_{1} \mathrm{n}_{1} \mathrm{~T}_{1}+\mathrm{F}_{2} \mathrm{n}_{2} \mathrm{~T}_{2}}{\mathrm{~F}_{1} \mathrm{n}_{1}+\mathrm{F}_{2} \mathrm{n}_{2}}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.