Question

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

Answer 1

Explanation:

Below is an attachment containing the solution.

Answer 2

Answer:

$P_{He}^|=288.85torr$

Explanation:

Given data

$P_{He}=750 torr\\V_{He}=285mL\\P_{Ar}=732 torr\\V_{Ar}=455 mL$

Required

Partial pressure of helium

Solution

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

So

$V_{total}=V_{He}+V_{Ar}\\V_{total}=285mL+455mL\\V_{total}=740mL$

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

So

$P_{He}V_{He}=P_{He}^|V_{total}\\P_{He}^|=\frac{P_{He}V_{He}}{V_{total}}\\P_{He}^|=\frac{750torr*285mL}{740mL}\\ P_{He}^|=288.85torr$