Question

One mole of Hydrogen gas at temperature $\text{'}\mathrm{t}\text{'}$ is mixed with one mole of Helium gas at temperature $\text{'}2\mathrm{t}\text{'}$ in an isolated vessel. The equilibrium temperature of the gas mixture is:

Answer 1

Step 1: Given data:

Mole of hydrogen gas present = 1

Mole of Helium gas present = 1

Temperature at which $1\mathrm{mole}$ of Hydrogen gas present = $\mathrm{t}$

Temperature at which $1\mathrm{mole}$ of Helium gas present = $2\mathrm{t}$

Step 2: Finding the value of equilibrium temperature of the gas mixture.

Because, ${\left({\mathrm{C}}_{\mathrm{v}}\right)}_{\mathrm{mix}}=\frac{{\mathrm{n}}_1{\mathrm{C}}_{\mathrm{v}1}+{\mathrm{n}}_2{\mathrm{C}}_{\mathrm{v}2}}{{\mathrm{n}}_1+{\mathrm{n}}_2}$

So, $$\begin{gathered} \frac{1\times \left(\frac{5\mathrm{R}}{2}\right)+1\times \left(\frac{3\mathrm{R}}{2}\right)}{1+1} \\ =2\mathrm{R} \end{gathered}$$

As, $\mathrm{U}={\mathrm{U}}_1+{\mathrm{U}}_2={\mathrm{n}}_1{\mathrm{C}}_{\mathrm{v}1}{\mathrm{T}}_1+{\mathrm{n}}_2{\mathrm{C}}_{\mathrm{v}2}{\mathrm{T}}_2$

Therefore: $({\mathrm{n}}_1+{\mathrm{n}}_2)\times {\left({\mathrm{C}}_{\mathrm{v}}\right)}_{\mathrm{mix}}\times {\mathrm{T}}_{\mathrm{m}}=1\times \frac{5\mathrm{R}}{2}\times \mathrm{T}+1\times \frac{3\mathrm{R}}{2}\times 2\mathrm{T}$.

as, ${\mathrm{n}}_1+{\mathrm{n}}_2=1+1=2$

So, for temperature: $$\begin{gathered} 2\times 2\mathrm{R}\times {\mathrm{T}}_{\mathrm{m}}=\frac{5\mathrm{RT}}{2}+3\mathrm{RT} \\ 4{\mathrm{RT}}_{\mathrm{m}}=\frac{11}{2}\mathrm{RT} \\ {\mathrm{T}}_{\mathrm{m}}=\frac{11}{8}\mathrm{T} \end{gathered}$$

Therefore, The equilibrium temperature of the gas mixture is: ${\mathrm{T}}_{\mathrm{m}}=\frac{11}{8}\mathrm{T}$