   Chapter 13, Problem 152CP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
3 views

# steel cylinder contains 150.0 moles of argon gas at a temperatute 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:

To calculate the mass of argon gas remains in the cylinder.

Concept Introduction:

The combine the gas equation gives the relation between pressure, volume and temperature at two different conditions.

PV= nRT.

Explanation

Given information:

P1 = 8.93 MPa = 88.1 atm (1 MPa = 9.86923 atm).

T1 = 25°C = 25 + 273 = 298 K.

n1 = 150.0 mole.

P2 = 2.00 MPa = 19.7 atm (1 MPa = 9.86923 atm).

T2 = 19°C = 19+273 = 292 K.

We need to calculate the initial volume with the help of given pressure, moles and temperature. Then we have to use that volume to calculate the moles of argon remains after removal of some gas. These moles will give the mass of argon remains in the cylinder.

P1V1= n1RT1

V1= n1RT1P1

Plug in the given values,

V1= n1RT1P1

V1=150

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