Question

A 5 L vessel contains 1.4 g of nitrogen gas $N_2$. When heated to 1800 K, 30% of molecules are dissociated into atoms. Calculate the pressure of the gas at 1800 K. [Mol. mass of $N_2=28mo{l}^{-1}$, R = 0.0821 L atm $K^{-1}mo{l}^{-1}$

• A. 2.45 atm
• B. 3.69 atm
• C. 1.92 atm
• D. None of these

Answer 1

The correct option is C 1.92 atm

Since some of the nitrogen gas is dissociated into atoms the vessel will now contains a mixture of molecular nitrogen and atomic nitrogen, Let partial pressure as $P_1$ & $P_2$

mass of molecular nitrogen $m_1=1.4\times \frac{70}{100}$

$=0.98g$

mass of atomic nitrogen $=m_2=1.4-0.98=0.42g$

acc to Dalton law,

total pressure $=p_1+p_2$

$p=p_1+p_2=\frac{n_{1}RT}{V}+\frac{n_{2}RT}{V}=(\frac{0.98}{28}\times \frac{0.0821\times 1800}{5})+(\frac{0.42}{14}\times \frac{0.0821\times 1800}{5})$

$P=1.0344+0.887=1.9214atm$