Question

Three perfect gases at absolute temperature ${{\text{T}}_{1}},{{\text{T}}_{2}}$ and${{\text{T}}_{3}}$ are mixed. The masses of molecules are${{\text{m}}_{1}},{{\text{m}}_{2}}$ and${{\text{m}}_{3}}$ and the number of molecules are${{\text{n}}_{1}},{{\text{n}}_{2}}$ and${{\text{n}}_{3}}$ respectively. Assuming no less of energy, the final temperature of the mixture is:
A.$\dfrac{\left( {{\text{T}}_{1}}+{{\text{T}}_{2}}+{{\text{T}}_{3}} \right)}{3}$
B.$\dfrac{{{\text{n}}_{1}}{{\text{T}}_{1}}+{{\text{n}}_{2}}{{\text{T}}_{2}}+{{\text{n}}_{3}}{{\text{T}}_{3}}}{{{\text{n}}_{1}}+{{\text{n}}_{2}}+{{\text{n}}_{3}}}$
C.$\dfrac{{{\text{n}}_{1}}{{\text{T}}_{1}}^{2}+{{\text{n}}_{2}}{{\text{T}}_{2}}^{2}+{{\text{n}}_{3}}{{\text{T}}_{3}}^{2}}{{{\text{n}}_{1}}{{\text{T}}_{1}}+{{\text{n}}_{2}}{{\text{T}}_{2}}+{{\text{n}}_{3}}{{\text{T}}_{3}}}$
D.$\dfrac{{{\text{n}}_{1}}^{2}{{\text{T}}_{1}}^{2}+{{\text{n}}_{2}}^{2}{{\text{T}}_{2}}^{2}+{{\text{n}}_{3}}^{2}{{\text{T}}_{3}}^{2}}{{{\text{n}}_{1}}{{\text{T}}_{1}}+{{\text{n}}_{2}}{{\text{T}}_{2}}+{{\text{n}}_{3}}{{\text{T}}_{3}}}$

Answer 1

The K.E of n molecules given by$\text{n}\left( \dfrac{1}{2}\text{ }{{\text{K}}_{\text{B}}}\text{T} \right)$. The total kinetic energy to the sum of kinetic energies of individual gases from this, the required result can be obtained.

Complete answer:
For perfect gases.
Kinetic energy of n molecules$\text{n}\left( \dfrac{1}{2}\text{ }{{\text{K}}_{\text{B}}}\text{T} \right)$
As there are three perfect gases.
So K.E$=\left( {{\text{n}}_{1}}+{{\text{n}}_{2}}+{{\text{n}}_{3}} \right)\left( \dfrac{1}{2}{{\text{K}}_{\text{B}}}\text{T} \right)$
Now, total kinetic energy is equal to the sum of kinetic energies of individual gases.
So, n total K.E
$\begin{aligned} & ={{\text{n}}_{1}}{{\left( \text{K}\text{.E} \right)}_{1}}+{{\text{n}}_{2}}{{\left( \text{K}\text{.E} \right)}_{2}}+{{\text{n}}_{3}}{{\left( \text{K}\text{.E} \right)}_{3}} \\\ & \left( {{\text{n}}_{1}}+{{\text{n}}_{2}}+{{\text{n}}_{3}} \right)\left( \dfrac{1}{2}\text{ }{{\text{K}}_{\text{B}}}\text{T} \right)={{\text{n}}_{1}}\left( \dfrac{1}{2}\text{ }{{\text{K}}_{\text{B}}}{{\text{T}}_{1}} \right)+{{\text{n}}_{2}}\left( \dfrac{1}{2}\text{ }{{\text{K}}_{\text{B}}}{{\text{T}}_{2}} \right)+{{\text{n}}_{3}}\left( \dfrac{1}{2}\text{ }{{\text{K}}_{\text{B}}}{{\text{T}}_{3}} \right) \\\ \end{aligned}$
Here ${{\text{T}}_{1}},{{\text{T}}_{2}},{{\text{T}}_{3}}$ are temperatures of individual gases.
So,$\left( {{\text{n}}_{1}},{{\text{n}}_{2}},{{\text{n}}_{3}} \right)\text{T}={{\text{n}}_{1}}{{\text{T}}_{1}}+{{\text{n}}_{2}}{{\text{T}}_{2}}+{{\text{n}}_{3}}{{\text{T}}_{3}}$
Or $\text{T}=\dfrac{{{\text{n}}_{1}}{{\text{T}}_{1}}+{{\text{n}}_{2}}{{\text{T}}_{2}}+{{\text{n}}_{3}}{{\text{T}}_{3}}}{{{\text{n}}_{1}}+{{\text{n}}_{2}}+{{\text{n}}_{3}}}$

So, the correct option is (B).

Note:
Perfect gas is also called an ideal gas. The ideal gas law PV = nRT relies on the assumptions.
The gas consists of a large number of molecules which are in random motion.
The volume of the molecules is negligibly small compared to volume occupied by gas.
No forces act on the molecules except during elastic collisions of negligible duration.