Question

A vessel contains an ideal gas which is pumped out at a constant rate $\frac{dV}{dt}=r$. Assume that the process takes place at constant temperature and the pressure inside the vessel depends on time as $P=P_0{e}^{\left(-\frac{rt}{V_0}\right)}$ , where $P_0$ and $V_0$ are initial pressure and volume of the vessel respectively. To find out the time taken by half the original gas to be pumped out, which of the following data is not required?

• A. Initial pressure inside the vessel
• B. Volume of the vessel
• C. Rate of pumping $r$
• D. Temperature of the gas

Answer 1

Answer: (A), (D)

The correct options are
A Initial pressure inside the vessel
D Temperature of the gas

Given, relation between pressure and time
$P=P_0{e}^{\left(-\frac{rt}{V_0}\right)}.....\left(1\right)$
When half of the gas is removed, $P$ reduces to $\frac{P_0}{2}$
$\therefore P=\frac{P_0}{2}.....\left(2\right)$
[because temperature and volume of vessel are constant, pressure $P\propto n$ (no. of moles of gas)]
Substituting $\left(2\right)$ in $\left(1\right)$, we get
$\frac{P_0}{2}=P_0{e}^{-\frac{rt}{V_0}}$
$\Rightarrow \ln 2=\frac{rt}{V_0}$
$\Rightarrow t=\frac{V_0\times \ln 2}{r}$
Hence, we only require volume of the vessel and the pumping rate.
Thus, both options (a) and (d) are correct answers.