Question

a gas is contained in a vessel of volume vo at a pressure po. if the gas is to be pumped out by a suction pump of stroke volume v them the number of moles of gas remained in the vessel after two stroke is

Answer 1

Answer:

Option A) is correct answer.

$\dfrac{P_0V_0^3}{RT(V_0+V)^2}$

Explanation:

Given that initial volume and pressure are $V_0\ and\ P_0$.

Stroke Volume of suction pump = V

As per ideal gas equation:

$PV = nRT$

Where P and V are pressure and volume respectively

n is number of moles of gas

R is Gas Constant and

T is temperature

NOTE: If T remains constant, then after a stroke of pump:

PV = $P_1V_1$

So, after first stroke of pump, we can say that:

$P_0V_0 = P_1 (V_0+V)$

Also $P_0V_0 = nRT$

$\Rightarrow P_0V_0 = nRT = P_1 (V_0+V)\\\Rightarrow P_1 = \dfrac{P_0V_0}{V_0+V} ..... (1)$

Now, for 2nd stroke of pump, pressure is $P_1$.

Using above formula:

$P_1V_0=P_2(V_0+V)\\\Rightarrow P_2 = \dfrac{P_1V_0}{V_0+V}$

Putting value of $P_1$ from equation (1):

$P_2 = \dfrac{P_0V_0 }{V_0+V} \times \dfrac{V_0 }{V_0+V}\\P_2 = \dfrac{P_0V_0^2 }{(V_0+V)^2} ...... (2)$

Using the formula

$P_2V_0 = nRT\\\Rightarrow n = \dfrac{P_2V_0}{RT}$

Putting the value of $P_2$ from equation (2):

$\Rightarrow n = \dfrac{P_0V_0^2}{(V_0+V)^2}\times \dfrac{V_0}{RT}\\\Rightarrow n = \dfrac{P_0V_0^3}{RT(V_0+V)^2}$

So, option A) is correct.

$n = \dfrac{P_0V_0^3}{RT(V_0+V)^2}$