Question

Two perfect gases at absolute temperatures ${{{T}}_{{1}}}$ and ${{{T}}_2}$ are mixed. There is no loss of energy. The temperature of mixture, if masses of molecules are ${{{m}}_{{1}}}$ and ${{{m}}_{{2}}}$ and the number of molecules in the gases are ${{{n}}_{{1}}}$ and ${{{n}}_2}$ respectively, is:
A) $\dfrac{{{{{T}}_{{1}}}{{ + }}{{{T}}_{{2}}}}}{{{2}}}$
B) $\dfrac{{{{{n}}_{{1}}}{{{T}}_{{1}}}{{ + }}{{{n}}_{{2}}}{{{T}}_{{2}}}}}{{{{{n}}_{{1}}}{{ + }}{{{n}}_{{2}}}}}$
C) $\dfrac{{{{3 }}{{{n}}_{{1}}}{{{T}}_{{1}}}{{ + 5 }}{{{n}}_{{2}}}{{{T}}_{{2}}}}}{{{{3 }}{{{n}}_{{1}}}{{ + 5 }}{{{n}}_{{2}}}}}$
D) $\sqrt {{{{T}}_{{1}}}{{ }}{{{T}}_{{2}}}}$

Answer 1

The absolute temperature of two perfect gases is given and the temperature of mixture is asked. It is also mentioned that there is no loss of energy. Kinetic theory of gas at temperature T is ${{nf \times }}\dfrac{{{1}}}{{{2}}}{{ k T}}$. Apply the same formula for both the perfect gases. And then add them in order to get the kinetic energy before mixing. Now, total kinetic energy of the gaseous mixture, ${{E = }}{{{n}}_{{1}}}{{f \times }}\dfrac{{{1}}}{{{2}}}{{kT + }}{{{n}}_{{2}}}{{f \times }}\dfrac{{{1}}}{{{2}}}{{kT}}$. Now equate them as there is no loss of energy during mixing.

Complete step by step solution:
Given: At absolute temperatures ${{{T}}_{{1}}}$ and ${{{T}}_2}$, two perfect gases are mixed.
Masses of molecules are ${{{m}}_{{1}}}$ and ${{{m}}_{{2}}}$
The number of molecules in the gases are ${{{n}}_{{1}}}$ and ${{{n}}_2}$
To find: Temperature of the mixture.
Let degree of freedom of each molecules be ${{f}}$
Let us consider that the molecules of the gas do not interact with each other
Kinetic energy of one gas at temperature ${{{T}}_{{1}}}$ is given by
$\Rightarrow {{{n}}_{{1}}}{{ f \times }}\dfrac{{{1}}}{{{2}}}{{ k }}{{{T}}_{{1}}}$
Kinetic energy of one gas at temperature ${{{T}}_2}$ is given by
$\Rightarrow {{{n}}_2}{{ f \times }}\dfrac{{{1}}}{{{2}}}{{ k }}{{{T}}_2}$
Now, the total kinetic energy of the gas before mixing is given by
$\Rightarrow {{{E}}_{{i}}}{{ = }}\dfrac{{{f}}}{{{2}}}{{ k}}\left( {{{{n}}_{{1}}}{{ }}{{{T}}_{{1}}}{{ + }}{{{n}}_{{2}}}{{ }}{{{T}}_{{2}}}} \right)$
Let us consider that after mixing of the gases, the temperature of the mixture is T.
Now, the total kinetic energy of the gaseous mixture is given by
$\Rightarrow {{{E}}_{{f}}}{{ = }}{{{n}}_{{1}}}{{f \times }}\dfrac{{{1}}}{{{2}}}{{kT + }}{{{n}}_{{2}}}{{f \times }}\dfrac{{{1}}}{{{2}}}{{kT}} \\\ \Rightarrow {{{E}}_{{f}}}{{ = (}}{{{n}}_{{1}}}{{ + }}{{{n}}_{{2}}}{{)}}\dfrac{{{f}}}{{{2}}}{{kT}}$
As there is no loss of energy (given), then
Total kinetic energy of the gas before mixing = Total kinetic energy of the gas after mixing
$\Rightarrow {{{E}}_{{f}}}{{ = }}{{{E}}_{{i}}}$
$\Rightarrow {{ }}\dfrac{{{f}}}{{{2}}}{{ k}}\left( {{{{n}}_{{1}}}{{ }}{{{T}}_{{1}}}{{ + }}{{{n}}_{{2}}}{{ }}{{{T}}_{{2}}}} \right){{ = }}\dfrac{{{f}}}{{{2}}}{{ kT(}}{{{n}}_{{1}}}{{ + }}{{{n}}_{{2}}}{{)}}$
On rearranging the terms, we get
$\Rightarrow {{T = }}\dfrac{{{{{n}}_{{1}}}{{{T}}_{{1}}}{{ + }}{{{n}}_{{2}}}{{{T}}_{{2}}}}}{{{{{n}}_{{1}}}{{ + }}{{{n}}_{{2}}}}}$
Thus, the temperature of mixture, ${{T = }}\dfrac{{{{{n}}_{{1}}}{{{T}}_{{1}}}{{ + }}{{{n}}_{{2}}}{{{T}}_{{2}}}}}{{{{{n}}_{{1}}}{{ + }}{{{n}}_{{2}}}}}$.

Therefore, option (B) is the right answer.

Note: A gas which obeys the ideal gas equation i.e., ${{PV = nRT}}$, at all temperatures and pressures is called an ideal gas or perfect gas. While deriving the ideal gas equation, the following two equations are made: First assumption: The size of the gas molecules is negligible small. Second assumption: There is no force of attraction amongst the molecules of gas.