   # A steel cylinder contains 150.0 moles of argon gas at a temperature of 25°C and a pressure of 8.93 MPa. After some argon has been used, the pressure is 2.00 MPa at a temperature of 19°C. What mass of argon remains in the cylinder? ### 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 144CWP
Textbook Problem
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## A steel cylinder contains 150.0 moles of argon gas at a temperature of 25°C and a pressure of 8.93 MPa. After some argon has been used, the pressure is 2.00 MPa at a temperature of 19°C. What mass of argon remains in the cylinder?

Interpretation Introduction

Interpretation:

The mass of argon remaining in the given cylinder is needed to be determined if the initial number of moles of argon is 150mol at 25oC and a pressure of 8.93MPa is used up to a pressure of 2MPa at 19oC .

Concept introduction:

• According to ideal gas law,

If the volume of gas is constant, then the pressure is directly proportional to temperature and number of moles.

PnT=RV , P,T and n are proportional.

Mathematically for a gas,

PinitialninitialTinitial=PfinalnfinalTfinal

• Mass of a substance from its number of moles is,

Number of moles×Molecularmass in grams=Mass

### Explanation of Solution

Explanation

The initial number of moles of argon is given as 150mol .

The initial pressure of argon is given as 8.93MPa .

The initial temperature of argon is given as 25oC=298K .

The final pressure of argon is given as 2MPa .

The final temperature of argon is given as 19oC=292K .

According to ideal gas law,

For a gas,

If the volume of gas is constant, then the pressure is directly proportional to temperature and number of moles.

PnT=RV , P,T and n are proportional.

Mathematically for a gas,

PinitialninitialTinitial=PfinalnfinalTfinal

Therefore,

8.93MPa150mol×298K=2MPanfinal×292K

Then, the final number of moles of argon (remaining) is,

nfinal=150mol×298K×2MPa8

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