Question

One mole of an ideal gas goes through the process $P=\frac{2V^2}{1+V^2}$. Then, the change in temperature (in $\text{K}$) of the gas when volume changes from $1{\text{m}}^3$ to $2{\text{m}}^3$ is

• A. $\frac{4}{5R}$
• B. $\frac{11}{5R}$
• C. $\frac{5}{2R}$
• D. $\frac{5}{3R}$

Answer 1

The correct option is B $\frac{11}{5R}$

Given:
Initial volume, $V_1=1{\text{m}}^3$
Final volume, $V_2=2{\text{m}}^3$
Number of moles, $n=1$
Given, $P=\frac{2V^2}{1+V^2}$
To find:
Change in temperature, $\Delta T=?$

From, $P=\frac{2V^2}{1+V^2}$
$\Rightarrow \frac{RT}{V}=\frac{2V^2}{1+V^2}$
$[\text{From}PV=nRT$ and $n=1]$
$\Rightarrow T=\frac{2V^3}{R(1+V^2)}$
At $V_1=1{\text{m}}^3,T_1=\frac{1}{R}$
and at $V_2=2{\text{m}}^3$, $T_2=\frac{16}{5R}$
$\therefore \Delta T=T_2-T_1=\frac{16}{5R}-\frac{1}{R}=\frac{11}{5R}\text{K}$