JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Heat & Thermodynamics

Two ideal polyatomic gages at temperatures

T_{1}

and

T_{2}

are mixed so that there is no loss of energy. If

F_{1}

and

F_{2}, m_{1}

and

m_{2}, n_{1}

and

n_{2}

be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is:

(A) $\frac{n _{1} T _{1} + n _{2} T _{2}}{n _{1} + n _{2}}$
(B) $\frac{n _{1} F _{1} T _{1} + n _{2} F _{2} T _{2}}{n _{1} F _{1} + n _{2} F _{2}}$ ✓ Correct
(C) $\frac{n _{1} F _{1} T _{1} + n _{2} F _{2} T _{2}}{F _{1} + F _{2}}$
(D) $\frac{n _{1} F _{1} T _{1} + n _{2} F _{2} T _{2}}{n _{1} + n _{2}}$

Answer: (B)

Solution

Mixing with no loss of energy means the internal energies of the two samples add to that of the mixture. For a gas of

n

molecules with

F

degrees of freedom at temperature

T

U = \frac{F}{2} nk T

, so

\frac{F _{1}}{2} n_{1} k T_{1} + \frac{F _{2}}{2} n_{2} k T_{2} = \frac{F _{1}}{2} n_{1} k T + \frac{F _{2}}{2} n_{2} k T

Solving for

T

T = \frac{n _{1} F _{1} T _{1} + n _{2} F _{2} T _{2}}{n _{1} F _{1} + n _{2} F _{2}}

The temperature of the mixture is

\frac{n _{1} F _{1} T _{1} + n _{2} F _{2} T _{2}}{n _{1} F _{1} + n _{2} F _{2}}

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Single correct option