Question

Boyle's, Charles' and Avogadro's results on the different state variables of a gas, can be combined into one "equation of state” known as the ideal gas equation, PV = nRT, where R = 8.514 J/mol-K. Consider the following situation.

A mixture of hydrogen and oxygen has volume $2000c{m}^3$, temperature 300 K, pressure 100 kPa and mass 0.76 gms. Calculate the masses of hydrogen and oxygen in the mixture.

• A. 0.12 gms; 0.64 gms
• B. 0.8gms; 0.68 gms
• C. 0.24 gms; 0.52 gms
• D. 0.16gms; 0.60 gms

Answer 1

The correct option is A

0.12 gms; 0.64 gms

Suppose there are $n_1$ moles of hydrogen and $n_2$ moles of oxygen in the mixture. The pressure of the mixture will be -

$P=\left(\frac{n_{1}RT}{V}\right)+\left(\frac{n_{2}RT}{V}\right)=(n_1+n_2)\frac{RT}{V}$

$\Rightarrow 100\times {10}^{3}Pa=(n_1+n_2)\times \left[\frac{\left(8.3J{K}^{-1}mo{l}^{-1}\right)\left(300K\right)}{200\times {10}^{-6}{m}^{-3}}\right]$

$\Rightarrow (n_1+n_2)=0.08mol$ .....(i)

The mass of the mixture is -

$(n_1\times 2gms/mol)+(n_2\times 32gms/mol)=0.76gms$

$\Rightarrow (n_1+16n_2)=0.38mol.$ ...(ii)

From (i) and (ii),

$n_1=0.06moland{n}_2=0.02mol$

The mass of hydrogen = $0.06\times 2gms=0.12gms$, and the mass of oxygen = $0.02\times 32gms=0.64gms$.