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

0.9 b a r

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4 b a r

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1.8 b a r

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3.6 b a r

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Solution

The correct option is C $1.8 b a r$
For $H_{2}$

$P_{1} V_{1} = P_{2} V_{2}$

$0.8 \times 0.5 = P_{2} \times 1$

$∴ P_{2} = 0.40$ bar

For $O_{2}$

$P_{1} V_{1} = P_{2} V_{2}$

$0.7 \times 2 = P_{2} \times 1$

$∴ P_{2} = 1.4$ bar

$∴ P_{T} = P_{H_{2}} + P_{O_{2}} = (0.4 + 1.4) b a r = 1.8 b a r$

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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 L of dioxygen at 0.7 bar are introduced in a 1 L vessel at 27∘C?

The correct option is C 1.8 barFor H2 P1V1=P2V2 0.8×0.5=P2×1 ∴P2=0.40 bar For O2 P1V1=P2V2 0.7×2=P2×1 ∴P2=1.4 bar ∴PT=PH2+PO2=(0.4+1.4) bar=1.8 bar

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