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  • Three perfect gases at absolute temperatures <img alt="T_{1},T_{2} \: and \: T_{3}" src="https://learn.careers360.com/latex-image/?%5Cdpi%7B100%7D%20T_%7B1%7D%2CT_%7B2%7D%20%5C%3A%20and%20%5C%3A%20

Three perfect gases at absolute temperatures T_{1},T_{2} \: and \: T_{3} are mixed. The masses of molecules are m_{1},m_{2} \: and \: m_{3} and the number of molecules are n_{1},n_{2} \: and \: n_{3} respectively. Assuming no loss of energy, the final temperature of the mixture is

Option: 1

\frac{\left ( T_{1}+T_{2}+T_{3} \right )}{3}


Option: 2

\frac{n_{1}T_{1}+n_{2}T_{2}+n_{3}T_{3} }{n_{1}+n_{2}+n_{3}}


Option: 3

\frac{n_{1}T_{1}^{2}+n_{2}T_{2}^{2}+n_{3}T_{3}^{2} }{n_{1}T_{1}+n_{2}T_{2}+n_{3}T_{3}}


Option: 4

\frac{n_{1}^{2}T_{1}^{2}+n_{2}^{2}T_{2}^{2}+n_{3}^{2}T_{3}^{2} }{n_{1}T_{1}+n_{2}T_{2}+n_{3}T_{3}}


Answers (1)

best_answer

For perfect gas,

Kinetic Energy of n molecule, K . E=n\left(\frac{1}{2} K_{B} T\right)$
Where, $K_{B}$ is Boltzmann constant
If there is no loss of energy.
Total kinetic energy of mixture is sum of each gas kinetic energy.

n_{\text {total}} K \cdot E_{\text {total}}=n_{1} K \cdot E_{1}+n_{2} K \cdot E_{2}+n_{3} K \cdot E_{3}$\\ $\left(n_{1}+n_{2}+n_{3}\right)\left(\frac{1}{2} K_{B} T\right)=n_{1}\left(\frac{1}{2} K_{B} T_{1}\right)+n_{2}\left(\frac{1}{2} K_{B} T_{2}\right)+n_{3}\left(\frac{1}{2} K_{B} T_{3}\right)$ $T=\frac{n_{1} T_{1}+n_{2} T_{2}+n_{3} T_{3}}{n_{1}+n_{2}+n_{3}}$

Posted by

Pankaj Sanodiya

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