Question

What will be the pressure of the gaseous mixture when $0.5\, L$ of $H_{2}$ at $0..8$ bar and $2.0 \,L$ of $O{2}$ at $0.7$ bar are introduced in a $1 \,L$ vessel at $27 \,{}^{\circ}C $?

• A. 1.8 bar
• B. 2.8 bar
• C. 3.0 bar
• D. 5 bar

Answer 1

Answer: (A)

Partial pressure of hydrogen gas
$V_{1}$ = $0.5 \,L$ $\quad$ $V_{2}$ = $1.0\,L$
$P_{1} =0.8 \,bar$ $\quad$ $P_{2} = ?$
Applying Boyle’s law $\left(constant \,T\, and \, n \right)$
$P_{1}V_{1}$ = $P_{2}V_{2}$ $\quad$ or $\quad$ $P_{2}$ =$\frac{P_{1}V_{1}}{V_{2}}$
$\therefore $ $\quad$ $P_{2}$ =$\frac{\left(0.8\,bar\right)\times\left(0.5\,L\right)}{\left(1.0\,L\right)}$ = $0.40\,bar$
Partial pressure of oxygen gas,
$V_{1}$ = $2 .0\, L$, $\quad$ $V_{2}$ = $1.0 \, L$
$P_{1}=0.7\, bar$ $\quad$ $P_{2} = ?$
or $P_{2}$ = $\frac{P_{1}V_{1}}{V_{2}}$ =$\frac{\left(0.7\,bar\right)\times\left(2.0\,L\right)}{\left(1.0\,L\right)}$ = $1.40\, bar$
Pressure of the gas mixture,
$P_{mix}$ = $P_{H_2}+P_{O_2}=0.40+1.40$ = $1.80\, bar$