Question

A vessel of volume $V_0$ contains an ideal gas at pressure $p_0$ and temperature T. Gas is continuously pumped out of this vessel at a constant volume -rate dV/dt = r keeping the temperature constant. The pressure of the gas being taken out equals the pressure inside the vessel. Find (a) the pressure of the gas as a function of time, (b) the time taken before half the original gas is pumped out.

Answer 1

We have,

$\frac{dV}{dt}$ = r

$\Rightarrow$ dV= rdt

Let the pressure pumped out gas = dp

Volume of container = $V_0$

At a pump dV amount of gas has been pumped out

$PdV=-V_{0}dP$

$\Rightarrow Prdt=-V_{0}dP$

$\Rightarrow \frac{dp}{P}=-\frac{rdt}{V_0}$

On integration, we get

P=e $\frac{rt}{V_0}$

P=e

Half of the gas been pumped out, pressure will be half

i.e. 1 = $\frac{1}{2}{e}^{\frac{rt}{V_0}}$

$\Rightarrow In2=\frac{rt}{V_0}$

$\Rightarrow t=In2\times \frac{V_0}{r}$