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  • Two ideal gases at absolute temperatures  and are mixed with each other. If the molecular mass and the number

Two ideal gases at absolute temperatures T_{1} and are mixed with each other. If the molecular mass and the number of molecules are m_{1}, n_{1} and m_{2}, n_{2}, respectively, Find out the temperature of the mixture

Option: 1

T=\frac{n_{1} T_{1}}{n_{1}+n_{2}}


Option: 2

T=\frac{n_{1} T_{1}+n_{2} T_{2}}{n_{1}+n_{2}}


Option: 3

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


Option: 4

T=\frac{n_{2}+T_{2}}{n_{2}}


Answers (1)

best_answer

The molecular kinetic energy of the first gas =n_{1} \cdot \frac{3}{2} k T_{1} and that of the second gas =n_{2} \cdot \frac{3}{2} k T_{2}.

So the net energy =\frac{3}{2} k\left(n_{1} T_{1}+n_{2} T_{2}\right).

The number of molecules in the mixture =n_{1}+n_{2}. Let the temperature of the mixture le T.

The tot molecular kinetic energy =\left(n_{1}+n_{2}\right) \cdot \frac{3}{2} k T

From the principle of energy conservation,

\left(n_{1}+n_{2}\right) \cdot \frac{3}{2} k T=\frac{3}{2} k\left(n_{1} T_{1}+n_{2} T_{2}\right)
or, T=\frac{n_{1} T_{1}+n_{2} T_{2}}{n_{1}+n_{2}}.
 

Posted by

Devendra Khairwa

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