Question

Suppose an ideal gas undergoes a process given by $V{P}^3=\text{constant}$. Initial temperature and volume of the gas are $T$ and $V$ respectively. If the gas expands to $27V$, then its temperature will become

• A. $T$
• B. $9T$
• C. $27T$
• D. $\frac{T}{9}$

Answer 1

The correct option is B $9T$

Given,$V{P}^3=\text{constant}=k^3$ (say) $\Rightarrow P=\frac{k}{V^{1/3}}......\left(1\right)$
Also for an ideal gas, $PV=nRT$
Using $\left(1\right)$ in the above ideal gas equation, we get
$\frac{k}{V^{1/3}}.V=nRT\Rightarrow V^{2/3}=\frac{nRT}{k}$
$\therefore$ we can say that, $V^{2/3}\propto T$
Hence, ${\left(\frac{V_1}{V_2}\right)}^{2/3}=\frac{T_1}{T_2}.....\left(2\right)$
Given, Initial temperature $\left(T_1\right)=T$
Initial volume $\left(V_1\right)=V$
Final volume $\left(V_2\right)=27V$
Final temperature $\left(T_2\right)=?$
Equation $\left(2\right)$ becomes
${\left(\frac{V}{27V}\right)}^{2/3}=\frac{T}{T_2}\Rightarrow T_2=9T$
Thus, option (b) is the correct answer.