(a) Against what external pressure must a gas sample expand from 50.0 mL to 219 mL to do 8.34 J of work? (b) To what volume will a 200.0-mL gas sample be compressed by an external pressure of 7.8 atm if 100.0 J of work are done on the system? (c) To what volume must a 200.0-mL gas sample expand against an external pressure of 7.8 atm to do 100.0 J of work?

Question

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Answer 1

Textbook Question

Chapter 10.3, Problem 2PPB

(a) Against what external pressure must a gas sample expand from 50.0 mL to 219 mL to do 8.34 J of work? (b) To what volume will a 200.0-mL gas sample be compressed by an external pressure of 7.8 atm if 100.0 J of work are done on the system? (c) To what volume must a 200.0-mL gas sample expand against an external pressure of 7.8 atm to do 100.0 J of work?

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The external pressure is $0.487 atm$ .

Explanation of Solution

To calculate the external pressure

Given data:

Initial volume = $50 ml$

Final volume = $219 ml$

Change in volume = $(219-50)ml=169 ml$

Work done = $8.34 J$

$w=-PΔV$

$8.34 J=-P(169 ml)P=0.0493×1 litre.atm101.3 JoulesP=0.487 atm$

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The volume is $73.5 ml$ .

Explanation of Solution

To calculate the volume

Given data:

Initial volume = $200 ml$

Work done = $100.0 J$

Pressure = $7.8 atm$

$w=-PΔV$

$100 J×1 litre.atm101.3 J=0.987 Litre.atm7.8 atm=0.981 litre.atmChange in volume = 0.9877.8 atm=0.1265 litresInitial volume-Final volume=0.1265 litres0.200 litres-Final volume=0.1265 litresFinal volume=0.0735 litres0.0735 litres×1000 ml/Litre=73.5 ml$

Final volume= $73.5 ml$

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The volume is $73.5 ml$

Explanation of Solution

To calculate the volume

Given data:

Initial volume = $200 ml$

Work done = $100.0 J$

Pressure = $7.8 atm$

$w=-PΔV$

$100 J×1 litre.atm101.3 J=0.987 Litre.atm7.8 atm=0.981 litre.atmChange in volume = 0.9877.8 atm=0.1265 litresInitial volume-Final volume=0.1265 litres0.200 litres-Final volume=0.1265 litresFinal volume=0.0735 litres0.0735 litres×1000 ml/Litre=73.5 ml$

Final volume= $73.5 ml$

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Answer 2

Textbook Question

Chapter 10.3, Problem 2PPB

(a) Against what external pressure must a gas sample expand from 50.0 mL to 219 mL to do 8.34 J of work? (b) To what volume will a 200.0-mL gas sample be compressed by an external pressure of 7.8 atm if 100.0 J of work are done on the system? (c) To what volume must a 200.0-mL gas sample expand against an external pressure of 7.8 atm to do 100.0 J of work?

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The external pressure is $0.487 atm$ .

Explanation of Solution

To calculate the external pressure

Given data:

Initial volume = $50 ml$

Final volume = $219 ml$

Change in volume = $(219-50)ml=169 ml$

Work done = $8.34 J$

$w=-PΔV$

$8.34 J=-P(169 ml)P=0.0493×1 litre.atm101.3 JoulesP=0.487 atm$

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The volume is $73.5 ml$ .

Explanation of Solution

To calculate the volume

Given data:

Initial volume = $200 ml$

Work done = $100.0 J$

Pressure = $7.8 atm$

$w=-PΔV$

$100 J×1 litre.atm101.3 J=0.987 Litre.atm7.8 atm=0.981 litre.atmChange in volume = 0.9877.8 atm=0.1265 litresInitial volume-Final volume=0.1265 litres0.200 litres-Final volume=0.1265 litresFinal volume=0.0735 litres0.0735 litres×1000 ml/Litre=73.5 ml$

Final volume= $73.5 ml$

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The volume is $73.5 ml$

Explanation of Solution

To calculate the volume

Given data:

Initial volume = $200 ml$

Work done = $100.0 J$

Pressure = $7.8 atm$

$w=-PΔV$

$100 J×1 litre.atm101.3 J=0.987 Litre.atm7.8 atm=0.981 litre.atmChange in volume = 0.9877.8 atm=0.1265 litresInitial volume-Final volume=0.1265 litres0.200 litres-Final volume=0.1265 litresFinal volume=0.0735 litres0.0735 litres×1000 ml/Litre=73.5 ml$

Final volume= $73.5 ml$

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Answer 3

Textbook Question

Chapter 10.3, Problem 2PPB

(a) Against what external pressure must a gas sample expand from 50.0 mL to 219 mL to do 8.34 J of work? (b) To what volume will a 200.0-mL gas sample be compressed by an external pressure of 7.8 atm if 100.0 J of work are done on the system? (c) To what volume must a 200.0-mL gas sample expand against an external pressure of 7.8 atm to do 100.0 J of work?

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The external pressure is $0.487 atm$ .

Explanation of Solution

To calculate the external pressure

Given data:

Initial volume = $50 ml$

Final volume = $219 ml$

Change in volume = $(219-50)ml=169 ml$

Work done = $8.34 J$

$w=-PΔV$

$8.34 J=-P(169 ml)P=0.0493×1 litre.atm101.3 JoulesP=0.487 atm$

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The volume is $73.5 ml$ .

Explanation of Solution

To calculate the volume

Given data:

Initial volume = $200 ml$

Work done = $100.0 J$

Pressure = $7.8 atm$

$w=-PΔV$

$100 J×1 litre.atm101.3 J=0.987 Litre.atm7.8 atm=0.981 litre.atmChange in volume = 0.9877.8 atm=0.1265 litresInitial volume-Final volume=0.1265 litres0.200 litres-Final volume=0.1265 litresFinal volume=0.0735 litres0.0735 litres×1000 ml/Litre=73.5 ml$

Final volume= $73.5 ml$

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The volume is $73.5 ml$

Explanation of Solution

To calculate the volume

Given data:

Initial volume = $200 ml$

Work done = $100.0 J$

Pressure = $7.8 atm$

$w=-PΔV$

$100 J×1 litre.atm101.3 J=0.987 Litre.atm7.8 atm=0.981 litre.atmChange in volume = 0.9877.8 atm=0.1265 litresInitial volume-Final volume=0.1265 litres0.200 litres-Final volume=0.1265 litresFinal volume=0.0735 litres0.0735 litres×1000 ml/Litre=73.5 ml$

Final volume= $73.5 ml$

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Answer 4

Textbook Question

Chapter 10.3, Problem 2PPB

(a) Against what external pressure must a gas sample expand from 50.0 mL to 219 mL to do 8.34 J of work? (b) To what volume will a 200.0-mL gas sample be compressed by an external pressure of 7.8 atm if 100.0 J of work are done on the system? (c) To what volume must a 200.0-mL gas sample expand against an external pressure of 7.8 atm to do 100.0 J of work?

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The external pressure is $0.487 atm$ .

Explanation of Solution

To calculate the external pressure

Given data:

Initial volume = $50 ml$

Final volume = $219 ml$

Change in volume = $(219-50)ml=169 ml$

Work done = $8.34 J$

$w=-PΔV$

$8.34 J=-P(169 ml)P=0.0493×1 litre.atm101.3 JoulesP=0.487 atm$

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The volume is $73.5 ml$ .

Explanation of Solution

To calculate the volume

Given data:

Initial volume = $200 ml$

Work done = $100.0 J$

Pressure = $7.8 atm$

$w=-PΔV$

$100 J×1 litre.atm101.3 J=0.987 Litre.atm7.8 atm=0.981 litre.atmChange in volume = 0.9877.8 atm=0.1265 litresInitial volume-Final volume=0.1265 litres0.200 litres-Final volume=0.1265 litresFinal volume=0.0735 litres0.0735 litres×1000 ml/Litre=73.5 ml$

Final volume= $73.5 ml$

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The volume is $73.5 ml$

Explanation of Solution

To calculate the volume

Given data:

Initial volume = $200 ml$

Work done = $100.0 J$

Pressure = $7.8 atm$

$w=-PΔV$

$100 J×1 litre.atm101.3 J=0.987 Litre.atm7.8 atm=0.981 litre.atmChange in volume = 0.9877.8 atm=0.1265 litresInitial volume-Final volume=0.1265 litres0.200 litres-Final volume=0.1265 litresFinal volume=0.0735 litres0.0735 litres×1000 ml/Litre=73.5 ml$

Final volume= $73.5 ml$

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The external pressure is $0.487 atm$ .

Explanation of Solution

To calculate the external pressure

Given data:

Initial volume = $50 ml$

Final volume = $219 ml$

Change in volume = $(219-50)ml=169 ml$

Work done = $8.34 J$

$w=-PΔV$

$8.34 J=-P(169 ml)P=0.0493×1 litre.atm101.3 JoulesP=0.487 atm$

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The volume is $73.5 ml$ .

Explanation of Solution

To calculate the volume

Given data:

Initial volume = $200 ml$

Work done = $100.0 J$

Pressure = $7.8 atm$

$w=-PΔV$

$100 J×1 litre.atm101.3 J=0.987 Litre.atm7.8 atm=0.981 litre.atmChange in volume = 0.9877.8 atm=0.1265 litresInitial volume-Final volume=0.1265 litres0.200 litres-Final volume=0.1265 litresFinal volume=0.0735 litres0.0735 litres×1000 ml/Litre=73.5 ml$

Final volume= $73.5 ml$

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The volume is $73.5 ml$

Explanation of Solution

To calculate the volume

Given data:

Initial volume = $200 ml$

Work done = $100.0 J$

Pressure = $7.8 atm$

$w=-PΔV$

$100 J×1 litre.atm101.3 J=0.987 Litre.atm7.8 atm=0.981 litre.atmChange in volume = 0.9877.8 atm=0.1265 litresInitial volume-Final volume=0.1265 litres0.200 litres-Final volume=0.1265 litresFinal volume=0.0735 litres0.0735 litres×1000 ml/Litre=73.5 ml$

Final volume= $73.5 ml$

Answer 8

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The external pressure is $0.487 atm$ .

Explanation of Solution

To calculate the external pressure

Given data:

Initial volume = $50 ml$

Final volume = $219 ml$

Change in volume = $(219-50)ml=169 ml$

Work done = $8.34 J$

$w=-PΔV$

$8.34 J=-P(169 ml)P=0.0493×1 litre.atm101.3 JoulesP=0.487 atm$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer 13

Answer to Problem 2PPB

The external pressure is $0.487 atm$ .

Answer 14

Explanation of Solution

To calculate the external pressure

Given data:

Initial volume = $50 ml$

Final volume = $219 ml$

Change in volume = $(219-50)ml=169 ml$

Work done = $8.34 J$

$w=-PΔV$

$8.34 J=-P(169 ml)P=0.0493×1 litre.atm101.3 JoulesP=0.487 atm$

Answer 15

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The volume is $73.5 ml$ .

Explanation of Solution

To calculate the volume

Given data:

Initial volume = $200 ml$

Work done = $100.0 J$

Pressure = $7.8 atm$

$w=-PΔV$

$100 J×1 litre.atm101.3 J=0.987 Litre.atm7.8 atm=0.981 litre.atmChange in volume = 0.9877.8 atm=0.1265 litresInitial volume-Final volume=0.1265 litres0.200 litres-Final volume=0.1265 litresFinal volume=0.0735 litres0.0735 litres×1000 ml/Litre=73.5 ml$

Final volume= $73.5 ml$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer 20

Answer to Problem 2PPB

The volume is $73.5 ml$ .

Answer 21

Explanation of Solution

To calculate the volume

Given data:

Initial volume = $200 ml$

Work done = $100.0 J$

Pressure = $7.8 atm$

$w=-PΔV$

$100 J×1 litre.atm101.3 J=0.987 Litre.atm7.8 atm=0.981 litre.atmChange in volume = 0.9877.8 atm=0.1265 litresInitial volume-Final volume=0.1265 litres0.200 litres-Final volume=0.1265 litresFinal volume=0.0735 litres0.0735 litres×1000 ml/Litre=73.5 ml$

Final volume= $73.5 ml$

Answer 22

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The volume is $73.5 ml$

Explanation of Solution

To calculate the volume

Given data:

Initial volume = $200 ml$

Work done = $100.0 J$

Pressure = $7.8 atm$

$w=-PΔV$

$100 J×1 litre.atm101.3 J=0.987 Litre.atm7.8 atm=0.981 litre.atmChange in volume = 0.9877.8 atm=0.1265 litresInitial volume-Final volume=0.1265 litres0.200 litres-Final volume=0.1265 litresFinal volume=0.0735 litres0.0735 litres×1000 ml/Litre=73.5 ml$

Final volume= $73.5 ml$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Interpretation Introduction

Interpretation:

The external pressure and volumes has to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$H=U+PV$

Where,

H=enthalpy

U=internal energy

P=Pressure

V=Volume

The work done on the system or by the system can be calculated from the equation

$w=-PΔV$

Where,

w=work done

P=Pressure

V=Change in volume

Answer 27

Answer to Problem 2PPB

The volume is $73.5 ml$

Answer 28

Explanation of Solution

To calculate the volume

Given data:

Initial volume = $200 ml$

Work done = $100.0 J$

Pressure = $7.8 atm$

$w=-PΔV$

$100 J×1 litre.atm101.3 J=0.987 Litre.atm7.8 atm=0.981 litre.atmChange in volume = 0.9877.8 atm=0.1265 litresInitial volume-Final volume=0.1265 litres0.200 litres-Final volume=0.1265 litresFinal volume=0.0735 litres0.0735 litres×1000 ml/Litre=73.5 ml$