Question

A rigid tank contains 1.50 moles of an ideal gas. Determine the number of moles of gas that must be withdrawn from the tank to lower the pressure of the gas from 25.0 atm to 5.00 atm. Assume the volume of the tank and the temperature of the gas remain constant during this operation.

Answer 1

Solution;

From Ideal gas laws;

$\frac{p_fv_f}{p_iv_i}=\frac{n_fRT_f}{n_iRT_i}$

From which we obtain;

$n_f=\frac{p_f}{p_i}n_i$

$n_f=\frac{5}{25}×1.5$ =0.3moles

So the number of moles to be withdrawn is;

$\Delta n=n_i-n_f=1.5-0.3=1.2moles$