Question

A closed vessel contains ozone, dinitrogen and dihydrogen gas and exerta
pressure of P(mmHg) if sample consists of equal number of atoms of oxygen
nitrogen and hydrogen then what will be the pressure when one is removed
*0.75P
*0.67P
*P
*1.33P​

Answer 1

Answer:

The pressure will be 0.67P.

Explanation:

From the ideal gas equation we have,

$P=\frac{nRT}{V}$       (1)

Where,

P=pressure of the gas

V=volume associated with the gas

n=number of moles of the gas

R=universal gas constant

T=temperature at which gas is present

By considering R, T, and V constant we can write equation (1) as,

$P=Kn$     (2)

K=constant

We can modify equation (2) as,

$\frac{P_1}{P_2} =\frac{n_1}{n_2}$       (3)

P₁=initial pressure

P₂=final pressure

n₁=total initial moles of the gases

n₂=final moles of the gases

From the question we have,

P₁=P

n₁=3n

n₂=2n     (as one of the gases is removed)

By substituting the required values in equation (3) we get;

$\frac{P}{P_2} =\frac{3n}{2n}$

$P_2=P\times \frac{2}{3}$

$P_2=0.67P$

Hence, 0.67P will be the pressure when one of the gas is removed.