Question

A vessel of volume 5 litre contains 1.4g of nitrogen at temperature 1800K. Find the pressure of the gas if 30% of its molecules are dissociated into atoms at this temperature.

Answer 1

Answer: The pressure of nitrogen gas after 30% dissociation is 1.91 atm.

Explanation:

              $N_2\rightleftharpoons 2N$

initial   100 %

eq'm    70%           30%

Mass $N_2$ gas at equilibrium = 70% of 1.4 g= $\frac{1.4\times 70}{100}=0.98 g$

Moles of undissociated $N_2$ at equilibrium =$\frac{\text{mass of }N_2}}{\text{molar mass of}N_2}=\frac{0.98 g}{28 g/mol}=0.035 mol$

Moles of dissociated $N_2$ at equilibrium=$\frac{\text{mass of }N_2}}{\text{molar mass of N_2}}=\frac{0.42 g}{28 g/mol}=0.015 mol$

According to reaction,1  mole nitrogen gas gives 2 atoms of Nitrogen.

0.015 mol nitrogen gas will give = $0.015 moles\times 2=0.03$ moles of Nitrogen atom.

Total moles of gases in container of 5L volume at 1800 K: 0.035+0.03 moles

PV=nRT

n = total moles of gases in container

$P=\frac{(0.035+0.03) moles\times 0.0820 Latm/K mol\times 1800 K}{5 L}=1.91 atm$

The total pressure = 1.91 atm

partial pressure nitrogen gas : mole fraction of nitrogen × total pressure

$\chi_{N_2}\times P=\frac{0.035}{0.035+0.03}\times 1.91 atm =1.02 atm$

The pressure of nitrogen gas after 30% dissociation is 1.02 atm.

Answer 2

Answer:

please refer attachment

Explanation: