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?

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,

$\text{H=U+PV}$

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

$\text{w=-PΔV}$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The external pressure is $\text{0}\text{.487}\text{atm}$ .

Explanation of Solution

To calculate the external pressure

Given data:

Initial volume = $\text{50}\text{ml}$

Final volume = $\text{219}\text{ml}$

Change in volume = $\text{(219-50)ml=169}\text{ml}$

Work done = $\text{8}\text{.34}\text{J}$

$\text{w=-PΔV}$

$\begin{array}{l}\text{8}\text{.34}\text{J=-P(169}\text{ml)}\\ \\ \text{P=0}\text{.0493×}\frac{\text{1}\text{litre}\text{.atm}}{\text{101}\text{.3}\text{Joules}}\\ \\ \text{P=0}\text{.487}\text{atm}\end{array}$

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,

$\text{H=U+PV}$

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

$\text{w=-PΔV}$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The volume is $\text{73}\text{.5}\text{ml}$ .

Explanation of Solution

To calculate the volume

Given data:

Initial volume = $\text{200 ml}$

Work done = $\text{100}\text{.0}\text{J}$

Pressure = $7.8\text{atm}$

$\text{w=-PΔV}$

$\begin{array}{l}\text{100}\text{J×}\frac{\text{1}\text{litre}\text{.atm}}{\text{101}\text{.3}\text{J}}\text{=0}\text{.987}\text{Litre}\text{.atm}\\ \\ \text{7}\text{.8}\text{atm=0}\text{.981}\text{litre}\text{.atm}\\ \\ \text{Change in volume = }\frac{\text{0}\text{.987}}{\text{7}\text{.8}\text{atm}}\text{=0}\text{.1265}\text{litres}\\ \\ \text{Initial}\text{volume-Final}\text{volume=0}\text{.1265}\text{litres}\\ \\ \text{0}\text{.200}\text{litres-Final}\text{volume=0}\text{.1265}\text{litres}\\ \\ \text{Final}\text{volume=0}\text{.0735}\text{litres}\\ \\ \text{0}\text{.0735}\text{litres×1000}\text{ml/Litre=73}\text{.5}\text{ml}\end{array}$

Final volume= $\text{73}\text{.5}\text{ml}$

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,

$\text{H=U+PV}$

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

$\text{w=-PΔV}$

Where,

w=work done

P=Pressure

V=Change in volume

Answer to Problem 2PPB

The volume is $\text{73}\text{.5}\text{ml}$

Explanation of Solution

To calculate the volume

Given data:

Initial volume = $\text{200 ml}$

Work done = $\text{100}\text{.0}\text{J}$

Pressure = $7.8\text{atm}$

$\text{w=-PΔV}$

$\begin{array}{l}\text{100}\text{J×}\frac{\text{1}\text{litre}\text{.atm}}{\text{101}\text{.3}\text{J}}\text{=0}\text{.987}\text{Litre}\text{.atm}\\ \\ \text{7}\text{.8}\text{atm=0}\text{.981}\text{litre}\text{.atm}\\ \\ \text{Change in volume = }\frac{\text{0}\text{.987}}{\text{7}\text{.8}\text{atm}}\text{=0}\text{.1265}\text{litres}\\ \\ \text{Initial}\text{volume-Final}\text{volume=0}\text{.1265}\text{litres}\\ \\ \text{0}\text{.200}\text{litres-Final}\text{volume=0}\text{.1265}\text{litres}\\ \\ \text{Final}\text{volume=0}\text{.0735}\text{litres}\\ \\ \text{0}\text{.0735}\text{litres×1000}\text{ml/Litre=73}\text{.5}\text{ml}\end{array}$

Final volume= $\text{73}\text{.5}\text{ml}$