Chapter 2, Problem 2.1PR

(a)

Interpretation Introduction

Interpretation:

An expression for work done for a gas obeying real gas equation where repulsion predominates has to interpret and also it has to be checked whether the work done is more or less than work done for the ideal gas.

Concept introduction:

Van der Waals equation:

Van der Waals equation represents the real gas equation.  Real gas molecules have their own volume and there is force of attraction and repulsion constantly working between the real gas molecules unlike the ideal gas molecules.  Hence ideal gas equation is subjected to modify with pressure and volume correction and thus van der Waals equation has been formed for real gases.

  $\left(\text{P}\text{+}\frac{{\text{an}}^{\text{2}}}{{\text{V}}^{\text{2}}}\right)\left(\text{V}\text{-}\text{nb}\right)\text{=}\text{nRT}$

Where,

P is pressure of the real gas

V is volume of the real gas

‘n’ is the number of moles of real gas molecules.

R is the gas constant

T is the temperature

‘a’ is the pressure correction term

‘b’ is the volume occupied by each molecules (volume correction term).

Ideal gas equation:

Ideal gases are those who have negligible amount of volume and no force of attraction and repulsion I between the gas molecules.  The ideal gas equation is given as,

  $\text{PV}\text{=}\text{nRT}$

Where,

P is pressure of the real gas

V is volume of the real gas

‘n’ is the number of moles of real gas molecules.

R is the gas constant

T is the temperature

Work done:

In thermodynamics work done by a system is to transfer energy from the system to surroundings. It is represented as,

$\text{dW}\text{=}\text{-PdV}$

$\begin{array}{l}\text{W}\text{=}\text{work}\text{done}\\ \\ \text{P}\text{=}\text{pressure}\text{of}\text{the}\text{system}\\ \\ \text{dV}\text{=}\text{change}\text{in}\text{the}\text{volume}\end{array}$

For expansion of volume work done is negative and for compression of volume work done is positive.

Explanation of Solution

From van der Waals equation we get,

  $\left(\text{P}\text{+}\frac{{\text{an}}^{\text{2}}}{{\text{V}}^{\text{2}}}\right)\left(\text{V}\text{-}\text{nb}\right)\text{=}\text{nRT}$

Now according to the given question repulsion term predominates and so van der Waals equation is reduced to,

  $\text{P}\text{=}\frac{\text{nRT}}{\left(\text{V}\text{-}\text{nb}\right)}$

So the work done by the gas following this equation of state has to be derived.

Applying the law of thermodynamics,

  $\text{dW}\text{=}\text{-PdV}$

Now replacing the value of P in the above equation from the equation of state and taking temperature constant it is obtained,

  $\text{dW}\text{=}\text{-}\frac{\text{nRT}}{\left(\text{V}\text{-}\text{nb}\right)}\text{dV}$

Now applying the integration for volume change says from ${\text{V}}_1$ to ${\text{V}}_2$ where ${\text{V}}_2\text{>}{\text{V}}_1$ and considering the change in work done as $\text{ΔW}$ it is found that,

  $\begin{array}{l}{\displaystyle \int \text{dW}=}-\text{nRT}{\displaystyle \underset{{\text{V}}_1}{\overset{{\text{V}}_2}{\int }}\frac{\text{dV}}{\left(\text{V}\text{-}\text{nb}\right)}}\\ \\ \text{ΔW}=-\text{nRT}\ln \left[\frac{{\text{V}}_2\text{-}\text{nb}}{{\text{V}}_1\text{-}\text{nb}}\right]\end{array}$

Now for ideal gas applying the ideal gas equation $\text{PV}\text{=}\text{nRT}$ in the equation of work done the work done by the ideal gas is obtained at constant temperature.

  $\text{dW}\text{=}\text{-PdV}$

  $\begin{array}{l}\text{P}\text{=}\frac{\text{nRT}}{\text{V}}\\ \\ \text{dW}\text{=}\text{-}\frac{\text{nRT}}{\text{V}}\text{dV}\\ \\ {\displaystyle \int \text{dW}\text{=}}\text{-}\text{nRT}{\displaystyle \underset{{\text{V}}_{\text{1}}}{\overset{{\text{V}}_{\text{2}}}{\int }}\frac{\text{dV}}{\text{V}}}\\ \\ \text{ΔW}\text{=}\text{-}\text{nRT}\text{ln}\left[\frac{{\text{V}}_{\text{2}}}{{\text{V}}_{\text{1}}}\right]\end{array}$

Now from the above equation it is found that ${\text{V}}_{\text{1}}\text{>}{\text{(V}}_{\text{1}}\text{-}\text{nb)}$ and ${\text{V}}_{\text{2}}\text{>}{\text{(V}}_{\text{2}}\text{-}\text{nb)}$ as $\text{b}$ is positive which means that $\text{|ΔW|}$ of real gas is less than $\text{|ΔW|}$ of ideal gas.

Because ideal gas has no attraction or repulsion in between the molecules but real gas has and here repulsion predominates.  So it is easier for real gas to expand as repulsion is more.

Hence work done by the ideal gas is more here.

(b)

Interpretation Introduction

Interpretation:

The expression of work done for a real gas where attraction force predominates has to be calculated and also whether the work done by the real gas is more or less than the work done by the ideal gas has to be interpreted also.

Concept introduction:

Van der Waals equation:

Van der Waals equation represents the real gas equation.  Real gas molecules have their own volume and there is force of attraction and repulsion constantly working between the real gas molecules unlike the ideal gas molecules.  Hence ideal gas equation is subjected to modify with pressure and volume correction and thus van der Waals equation has been formed for real gases.

  $\left(\text{P}\text{+}\frac{{\text{an}}^{\text{2}}}{{\text{V}}^{\text{2}}}\right)\left(\text{V}\text{-}\text{nb}\right)\text{=}\text{nRT}$

Where,

P is pressure of the real gas

V is volume of the real gas

‘n’ is the number of moles of real gas molecules.

R is the gas constant

T is the temperature

‘a’ is the pressure correction term

‘b’ is the volume occupied by each molecules (volume correction term).

Ideal gas equation:

Ideal gases are those who have negligible amount of volume and no force of attraction and repulsion I between the gas molecules. The ideal gas equation is given as,

  $\text{PV}\text{=}\text{nRT}$

Where,

P is pressure of the real gas

V is volume of the real gas

‘n’ is the number of moles of real gas molecules.

R is the gas constant

T is the temperature

Work done:

In thermodynamics work done by a system is to transfer energy from the system to surroundings. It is represented as,

$\text{dW}\text{=}\text{-PdV}$

Where,

W is the work done

P is the pressure

$\text{dV}$ is the change in volume.

For expansion of volume work done is negative and for compression of volume work done is positive.

Explanation of Solution

According to the given question in the real gas the attraction force predominates and so the van der Waals equation remains as it is.

  $\left(\text{P}\text{+}\frac{{\text{an}}^{\text{2}}}{{\text{V}}^{\text{2}}}\right)\left(\text{V}\text{-}\text{nb}\right)\text{=}\text{nRT}$

From this the expression for pressure is determined as,

  $\text{P}\text{=}\frac{\text{nRT}}{\left(\text{V}\text{-}\text{nb}\right)}\text{-}\frac{{\text{an}}^{\text{2}}}{{\text{V}}^{\text{2}}}$

From the expression of work, it is found that,

  $\text{dW}\text{=}\text{-PdV}$

Now to calculate the work done by the real gas at constant temperature for volume change say from ${\text{V}}_1$ to ${\text{V}}_2$ where ${\text{V}}_2\text{>}{\text{V}}_1$ and considering the change in work done as $\text{ΔW}$ the expression of pressure has to be substituted in the expression of work and it is found that,

  $\begin{array}{l}{\displaystyle \int \text{dW}}\text{=}-{\displaystyle \underset{{\text{V}}_{\text{1}}}{\overset{{\text{V}}_{\text{2}}}{\int }}\left(\frac{\text{nRT}}{\left(\text{V}\text{-}\text{nb}\right)}\text{-}\frac{{\text{an}}^{\text{2}}}{{\text{V}}^{\text{2}}}\right)}\text{dV}\\ \\ {\displaystyle \int \text{dW}}\text{=}-\text{nRT}{\displaystyle \underset{{\text{V}}_{\text{1}}}{\overset{{\text{V}}_{\text{2}}}{\int }}\left(\frac{\text{dV}}{\left(\text{V}\text{-}\text{nb}\right)}\right)}\text{+}{\text{an}}^{\text{2}}{\displaystyle \underset{{\text{V}}_{\text{1}}}{\overset{{\text{V}}_{\text{2}}}{\int }}\left(\frac{\text{dV}}{{\text{V}}^{\text{2}}}\right)}\\ \\ \text{ΔW}\text{=}-\text{nRT}\text{ln}\left(\frac{{\text{V}}_{\text{2}}\text{-}\text{nb}}{{\text{V}}_{\text{1}}\text{-}\text{nb}}\right)-{\text{an}}^{\text{2}}\left(\frac{\text{1}}{{\text{V}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{V}}_{\text{1}}}\right)\end{array}$

Now for ideal gas applying the ideal gas equation $\text{PV}\text{=}\text{nRT}$ in the equation of work done the work done by the ideal gas is obtained at constant temperature.

  $\text{dW}\text{=}\text{-PdV}$

  $\begin{array}{l}\text{P}\text{=}\frac{\text{nRT}}{\text{V}}\\ \\ \text{dW}\text{=}\text{-}\frac{\text{nRT}}{\text{V}}\text{dV}\\ \\ {\displaystyle \int \text{dW}\text{=}}\text{-}\text{nRT}{\displaystyle \underset{{\text{V}}_{\text{1}}}{\overset{{\text{V}}_{\text{2}}}{\int }}\frac{\text{dV}}{\text{V}}}\\ \\ \text{ΔW}\text{=}\text{-}\text{nRT}\text{ln}\left[\frac{{\text{V}}_{\text{2}}}{{\text{V}}_{\text{1}}}\right]\end{array}$

The repulsion term $\text{a}$ is a positive term and ${\text{V}}_2\text{>}{\text{V}}_1$ so the 2^nd term in the expression of work done by real gas becomes negative and considering the sign it will be positive.

Now the 1^st term in the expression of work done by real gas is less than the work done by the ideal gas as ${\text{V}}_{\text{1}}\text{>}{\text{(V}}_{\text{1}}\text{-}\text{nb)}$ and ${\text{V}}_{\text{2}}\text{>}{\text{(V}}_{\text{2}}\text{-}\text{nb)}$ as $\text{b}$ is positive if sign is not considered.

But anyway any logarithmic term will have less value than normal fraction and so for real gas the 2^nd term will predominate and hence work done by real gas will be more than ideal gas.

Because ideal gas has no attraction or repulsion in between the molecules but real gas has and here the attraction term predominates.  Now for expansion of real gas this attraction force will be a barrier and to remove this barrier more work has to be done by real gas for expansion.

So work done by the real gas having more attraction will be more than that of ideal gas.