What will be the pressure of the gaseous mixture when $0.5 L$ of $H_{2}$ at $0.8 b a r$ and $2.0 L$ of $O_{2}$ at $0.7 b a r$ are introduced in a $1 L$ vessel at $27^{\circ} C$ ?

1.8 b a r

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2.8 b a r

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3.0 b a r

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5 b a r

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Solution

The correct option is A $1.8 b a r$
Partial pressure of hydrogen gas
$V_{1} = 0.5 L V_{2} = 1.0 L$
$P_{1} = 0.8 b a r P_{2} = ?$
Applying Boyle's law (constant $T$ and $n$ )
$P_{1} V_{1} = P_{2} V_{2}$ or $P_{2} = \frac{P_{1} V_{1}}{V_{2}}$
$∴ P_{2} = \frac{(0.8 b a r) \times (0.5 L)}{(1.0 L)} = 0.40 b a r$
Partial pressure of oxygen gas,
$V_{1} = 2.0 L, V_{1} = 1.0 L$
$P_{1} = 0.7 b a r P_{2} = ?$
or $P_{2} = \frac{P_{1} V_{1}}{V_{2}} = \frac{(0.7 b a r) \times (2.0 L)}{(1.0 L)} = 1.40 b a r$ .
Pressure of the gas mixture,
$P_{m i n x} = P_{H_{2}} + P_{O_{2}} = 0.40 + 1.40 = 1.80 b a r$ .

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What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of O2 at 0.7 bar are introduced in a 1 L vessel at 27∘C?

What will be the pressure of the gaseous mixture when $0.5 L$ of $H_{2}$ at $0.8 b a r$ and $2.0 L$ of $O_{2}$ at $0.7 b a r$ are introduced in a $1 L$ vessel at $27^{\circ} C$ ?