Chapter 10, Problem 10.20QP

(a)

Interpretation Introduction

Interpretation:

The work done for the expansion of given gas should be calculated under given set of conditions.

Concept Introduction:

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

Enthalpy can be given by the formula,

$\begin{array}{l}\text{H=U+PV}\\ \begin{array}{c}\begin{array}{l}\text{Where,}\\ \text{ H=enthalpy}\end{array}\\ \text{U=internal energy}\\ \text{P=Pressure}\\ \text{V=Volume}\end{array}\end{array}$

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

$\begin{array}{l}\text{w=-PΔV}\\ \begin{array}{c}\begin{array}{l}\text{Where }\\ \text{w = work done}\end{array}\\ \text{P =}\text{Pressure}\\ \text{V=Change in volume}\end{array}\end{array}$

Answer to Problem 10.20QP

Work done is $\text{0J}$.

Explanation of Solution

Given,

Record the given info

Initial volume = $\text{1}\text{.6}\text{L}$

Final volume = $\text{5}\text{.4L}$

Calculate the work done against vacuum

$\text{w = -PΔV}$

$\begin{array}{l}\text{w = -(0)(5}\text{.4 - 1}\text{.6)L}\\ \text{w = 0 J}\end{array}$

Work done against vacuum is found to be $\text{0J}$

(b)

Interpretation Introduction

Interpretation:

The work done for the expansion of given gas should be calculated under given set of conditions.

Concept Introduction:

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

Enthalpy can be given by the formula,

$\begin{array}{l}\text{H=U+PV}\\ \begin{array}{c}\begin{array}{l}\text{Where,}\\ \text{ H=enthalpy}\end{array}\\ \text{U=internal energy}\\ \text{P=Pressure}\\ \text{V=Volume}\end{array}\end{array}$

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

$\begin{array}{l}\text{w=-PΔV}\\ \begin{array}{c}\begin{array}{l}\text{Where }\\ \text{w = work done}\end{array}\\ \text{P =}\text{Pressure}\\ \text{V=Change in volume}\end{array}\end{array}$

Answer to Problem 10.20QP

Work done is $\text{-3}{\text{.08×10}}^{\text{2}}\text{J}$.

Explanation of Solution

Given,

Record the given info

Initial volume = $\text{1}\text{.6}\text{L}$

Final volume = $\text{5}\text{.4L}$

Pressure = $\text{0}\text{.80}\text{atm}$

Calculate the work done against a constant pressure of $\text{0}\text{.80}\text{atm}$

$\begin{array}{l}\text{w=-PΔV}\\ \\ \text{w=-(0}\text{.80}\text{atm)(5}\text{.4-1}\text{.6)L}\\ \\ \text{w=-3}\text{.04}\text{L}\text{atm}\\ \\ \text{Converting}\text{to}\text{joules}\\ \\ \text{w=-3}\text{.04L}\text{.atm×}\frac{\text{101}\text{.3}\text{J}}{\text{1}\text{L}\text{.atm}}\\ \\ \text{w=-3}{\text{.08×10}}^{\text{2}}\text{J}\end{array}$

Therefore, work done at given constant pressure =$\text{-3}{\text{.08×10}}^{\text{2}}\text{J}$

c)

Interpretation Introduction

Interpretation:

The work done for the expansion of given gas should be calculated under given set of conditions.

Concept Introduction:

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

Enthalpy can be given by the formula,

$\begin{array}{l}\text{H=U+PV}\\ \begin{array}{c}\begin{array}{l}\text{Where,}\\ \text{ H=enthalpy}\end{array}\\ \text{U=internal energy}\\ \text{P=Pressure}\\ \text{V=Volume}\end{array}\end{array}$

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

$\begin{array}{l}\text{w=-PΔV}\\ \begin{array}{c}\begin{array}{l}\text{Where }\\ \text{w = work done}\end{array}\\ \text{P =}\text{Pressure}\\ \text{V=Change in volume}\end{array}\end{array}$

Answer to Problem 10.20QP

Work done is $\text{-1}{\text{.4×10}}^3\text{J}$

Explanation of Solution

Given,

Record the given info

Initial volume = $\text{1}\text{.6}\text{L}$

Final volume = $\text{5}\text{.4L}$

Pressure = $\text{3}\text{.7}\text{atm}$

Calculate the work done against a constant pressure of $\text{3}\text{.7}\text{atm}$

$\begin{array}{l}\text{w=-PΔV}\\ \\ \text{w=-(3}\text{.7}\text{atm)(5}\text{.4-1}\text{.6)L}\\ \\ \text{w=-14}\text{.06}\text{L}\text{atm}\\ \\ \text{Converting}\text{to}\text{joules}\\ \\ \text{w=-14}\text{.06L}\text{.atm×}\frac{\text{101}\text{.3}\text{J}}{\text{1}\text{L}\text{.atm}}\\ \\ \text{w=-1}{\text{.4×10}}^3\text{J}\end{array}$

Therefore, work done at given constant pressure =$\text{-1}{\text{.4×10}}^3\text{J}$