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Byju's Answer

Standard XII

Chemistry

Charles' Law

125 mL of a g...

Question

$125 m L$ of a gas $A$ of pressure $500 m m$ is mixed with $200 m L$ of another gas $B$ at a pressure of $300 m m$ in a vessel of $150 m L$ capacity. What will be the total pressure of the resulting mixture if the temperature is kept constant?

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Solution

Applying mole conservation,
$P_{t o t a l} = \frac{P_{1} V_{1} + P_{2} V_{2}}{V_{v e s s e l}}$
$= \frac{125 \times 500 + 200 \times 300}{150} = 816.67 m m$

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125 mL of a gas A of pressure 500 mm is mixed with 200 mL of another gas B at a pressure of 300 mm in a vessel of 150 mL capacity. What will be the total pressure of the resulting mixture if the temperature is kept constant?

Applying mole conservation,Ptotal=P1V1+P2V2Vvessel=125×500+200×300150=816.67 mm

$125 m L$ of a gas $A$ of pressure $500 m m$ is mixed with $200 m L$ of another gas $B$ at a pressure of $300 m m$ in a vessel of $150 m L$ capacity. What will be the total pressure of the resulting mixture if the temperature is kept constant?

Applying mole conservation,
$P_{t o t a l} = \frac{P_{1} V_{1} + P_{2} V_{2}}{V_{v e s s e l}}$
$= \frac{125 \times 500 + 200 \times 300}{150} = 816.67 m m$