Question

A vessel of volume $2{\text{m}}^3$ contains an ideal gas at pressure ${10}^5{\text{N/m}}^2$ and temperature $300\text{K}$. The gas is continously pumped out of this vessel at a constant volume rate $\frac{dV}{dt}=2\times {10}^{-3}{\text{m}}^3/\text{s}$ , keeping the temperature constant. The pressure of the gas being taken out equals the pressure inside the vessel. Find the time taken before half of the original gas is pumped out (in minutes).

• A. $5.5\text{min}$
• B. $11.55\text{min}$
• C. $10.5\text{min}$
• D. $21.5\text{min}$

Answer 1

The correct option is B $11.55\text{min}$

For the situation given above, the pressure variation is given by
$P=P_0{e}^{\left(-\frac{rt}{V_0}\right)}......\left(1\right)$
where $r=2\times {10}^{-3}{\text{m}}^3/\text{s}$
Pressure after pumping half of the gas out,
$P=\frac{P_0}{2}....\left(2\right)$
[$\because T\&V$ are constant, pressure $P\propto n$ and hence when half of the quantity of gas is taken out, pressure also becomes half]
Substituting $\left(2\right)$ in $\left(1\right)$, we get
$\frac{P_0}{2}=P_0{e}^{\left(-\frac{rt}{V_0}\right)}$
$\Rightarrow \ln 2=\frac{rt}{V_0}$
$\Rightarrow t=\frac{V_0\times \ln 2}{r}$
From the data given in the question,
$t=\frac{2\times \ln 2}{2\times {10}^{-3}}$
$=1000\times \ln 2$
$=693.1475\text{s}=11.55\text{min}(\because \ln 2=0.693)$
Thus, option (b) is the correct answer.