Question

A 14.0 L container at 323 K holds a mixture of two gases with a total pressure of 8.00 atm. If there are 1.50 mol of Gas A in the mixture, how many moles of Gas B are present

Answer 1

The number of moles of gas B present in the gas mixture is 2.72 moles.

The given parameters;

• Volume of the gases, V = 14 L
• Temperature of the gases, T = 323 K
• Total pressure of the gases, P = 8 atm
• Number of moles of gas A = 1.5 moles

The total number of moles of the gases is calculated by applying ideal gas law;

PV = nRT

where;

R is ideal gas constant = 0.0821 L.atm./mol.K

n is the total mole of the gases

$n =\frac{PV}{RT} \\\\n = \frac{8 \times 14}{0.0821 \times 323} \\\\n = 4.22 \ moles$

The number of moles of gas B is calculated as follows;

$n_B = 4.22 - n_A\\\\n_B = 4.22 - 1.5\\\\n_B = 2.72 \ moles$

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