Question

$n_1$ molecules of a gas is kept in a container of definite volume at a temperature of $T^{\circ }$ and at a pressure P. What will be the pressure of the gas at that temperature when $n_2$ molecules of gas is removed from the container?

• A. $\frac{n_{1}P}{(n_1-n_2)}$
• B. $\frac{(n_1-n_2)P}{n_2}$
• C. $\frac{(n_1-n_2)}{n_{2}P}$
• D. $\frac{(n_1-n_2)P}{n_1}$

Answer 1

The correct option is D $\frac{(n_1-n_2)P}{n_1}$

As per the ideal gas law; PV = nRT ; where P = pressure, V = volume, n = no. of moles, T = temperature & R = gas constant.
Therefore V = $\frac{n_{1}RT}{P}$
Also the new pressure when $n_2$ moles are removed = $\frac{(n_1-n_2)P}{n_1}$