   Chapter 5, Problem 28PS

Chapter
Section
Textbook Problem

As the gas trapped in a cylinder with a movable piston cools, 1.34 kJ of work is done on the gas by the surroundings. If the gas is at a constant pressure of 1.33 × 105 Pa, what is the change of volume (in L) of the gas?

Interpretation Introduction

Interpretation:

The change in volume under a constant pressure has to be determined.

Concept Introduction:

The amount work done to the system or from the system.

W=-P(ΔV)

Where,

ΔV is the change in volume

P is the pressure.

Explanation

Given,

Pressure=1.33×105P

Work done at constant pressure= 1.34kJ

Rearrange and substitute in above equation as,

ΔV=-W/P

ΔV=-1.34kJ/1.33×105Pa

ΔV=-1

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