   # Consider the three flasks in the diagram below. Assuming the connecting tubes have negligible volume, what is the partial pressure of each gas and the total pressure after all the stopcocks are opened? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 8, Problem 87E
Textbook Problem
6370 views

## Consider the three flasks in the diagram below. Assuming the connecting tubes have negligible volume, what is the partial pressure of each gas and the total pressure after all the stopcocks are opened? Interpretation Introduction

Interpretation: The total pressure, the partial pressures of each gas in the given figure is needed to be determined when the stop-cocks are opened.

Concept introduction:

• Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.
• Total pressure of a mixture of gases is the sum of individual partial pressures of constituted gases.
• According to Boyle’s law of gas molecules,

If the volume of gases in a container is increased then the pressure will decrease at constant temperature and number of moles.

Equation for Boyle’s law is,

PinitialVinitial=PfinalVfinal

### Explanation of Solution

Explanation

In the figure,

In initial condition, the stop-cocks of each container are closed.

In final condition, the stop-cocks of each container are opened.

Therefore,

The initial pressure of He gas is 200torr

The initial volume of He gas is 1L

The final volume is the sum of volumes of three containers =1L+1L+2L=4L

The partial pressure of He is the final pressure in according to Boyle’s law.

Equation for Boyle’s law states,

PinitialVinitial=PfinalVfinal

Therefore,

Partial pressure of He is,

Pfinal=200torr×1L4L =50torr

Hence, the partial pressure of He when the stop-cocks are opened in the figure is 50torr .

In the figure,

In initial condition, the stop-cocks of each container are closed.

In final condition, the stop-cocks of each container are opened.

Therefore,

The initial pressure of Ne gas is 0.4atm=0.4atm×760torr1atm=304torr

The initial volume of Ne gas is 1L

The final volume is the sum of volumes of three containers =1L+1L+2L=4L

The partial pressure of Ne is the final pressure in according to Boyle’s law.

Equation for Boyle’s law states,

PinitialVinitial=PfinalVfinal

Therefore,

Partial pressure of He is,

Pfinal=304torr×1L4L =76torr

Hence, the partial pressure of Ne when the stop-cocks are opened in the figure is 76torr

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