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Relating Translational KE and T

Two perfect g...

Question

Two perfect gases at absolute temperatures $T_{1}$ and $T_{2}$ are mixed. Assuming that there is no loss of energy. Find the temperature T of the mixture if the masses of the molecules are $m_{1}$ and $m_{1}$ and the number of molecules in the gases are $n_{1}$ and $n_{2}$ .

$T_{1} T_{2}$

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$\frac{T_{1}}{T_{2}}$

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

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

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Solution

The correct option is C
$\frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

According to the kinetic theory of gases, the kinetic energy of an ideal gas molecule at temperature $T$ is given by K.E = $\frac{3}{2} k T$ . As there is no force of attraction among the molecules of a perfect gas, PE. of the molecules is zero. So the energy of a molecule of a perfect gas is as follows:

E = K.E. + P.E. = $\frac{3}{2} k T + 0 = \frac{3}{2} k T$

Now, if T is the temperature of the mixture , by conservation of energy , we have

$n_{1} E_{1} + n_{2} E_{2} = (n_{1} + n_{2}) E$

or

$n_{1} (\frac{3}{2} k T_{1}) + n_{2} (\frac{3}{2} k T_{2}) = (n_{1} + n_{2}) \frac{3}{2} k T \Rightarrow T = \frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

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