Question

16 grams of $O_2$, 56 grams of $N_2$ & 4 grams of $H_2$ are mixed in a container of capacity 4.5 litres. If mixture is kept at $127{}^{\circ }C$, the pressure of container will be approximately equal to

• A. $33\times {10}^{5}N/m^2$
• B. $20\times {10}^{5}N/m^2$
• C.
  $14\times {10}^{5}N/m^2$
• D. $26\times {10}^{5}N/m^2$

Answer 1

The correct option is A $33\times {10}^{5}N/m^2$

$n_{O_2}=\frac{m}{M}=\frac{16}{32}=0.5n_{N_2}=\frac{m}{M}=\frac{56}{28}=2.0n_{H_2}=\frac{m}{M}=\frac{4}{2}=2.0n=n_1+n_2+n_3=4.5P=\frac{nRT}{V}=\frac{4.\text{/}5\times 8.3\times 400}{\text{/}4.5\times {10}^{-3}}=8.3\times 4\times {10}^5$ $\simeq 33\times {10}^5$