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A 285-mL flask contains pure helium at a pressure of 750 torr . A second flask with a volume of 455 mL contains pure argon at a pressure of 732 torr .If we connect the two flasks through a stopcock and we open the stopcock, what is the partial pressure of helium

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Khoso123

Answer:

[tex]P_{He}^|=288.85torr[/tex]

Explanation:

Given data

[tex]P_{He}=750 torr\\V_{He}=285mL\\P_{Ar}=732 torr\\V_{Ar}=455 mL[/tex]

Required

Partial pressure of helium

Solution

First calculate the total volume of gas mixture once the stopcock is opened

So

[tex]V_{total}=V_{He}+V_{Ar}\\V_{total}=285mL+455mL\\V_{total}=740mL[/tex]

As temperature remains constant,by Boyle's Law we calculate the partial pressure of the helium gas

So

[tex]P_{He}V_{He}=P_{He}^|V_{total}\\P_{He}^|=\frac{P_{He}V_{He}}{V_{total}}\\P_{He}^|=\frac{750torr*285mL}{740mL}\\ P_{He}^|=288.85torr[/tex]

Explanation:

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