   Chapter 5, Problem 30PS

Chapter
Section
Textbook Problem

The internal energy of a gas decreases by 1.65 kJ when it transfers 1.87 kJ of energy in the form of heat to the surroundings.(a) Calculate the work done by the gas on the surroundings.(b) Does the volume of gas increase or decrease?

(a)

Interpretation Introduction

Interpretation:

The work done by the gas has to be determined.

Concept Introduction:

The internal energy, U is the sum of kinetic energy and the potential energy inside the system.

Change in internal energy,ΔU, is the measures of energy that is transferred as heat and work from the system or to the system.

ΔU=qp+Wp

Where q = heat energy absorbed or released, W =work done to the system or from the system.

W=-P(ΔV)

Where,ΔV is the change in volume, and P is the pressure.

Explanation

Given,

Energy,q= 1.87kJ

ΔU=1.65kJ

Change in internal energy, ΔU=qp+Wp

(b)

Interpretation Introduction

Interpretation:

The volume of gas has to be determined.

Concept Introduction:

The internal energy, U is the sum of kinetic energy and the potential energy inside the system.

Change in internal energy,ΔU, is the measures of energy that is transferred as heat and work from the system or to the system.

ΔU=qp+Wp

Where q = heat energy absorbed or released, W =work done to the system or from the system.

W=-P(ΔV)

Where,ΔV is the change in volume, and P is the pressure.

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