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Expert-verified Found in: Page 9 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # Even at low density, real gases don’t quite obey the ideal gas law. A systematic way to account for deviations from ideal behavior is the virialexpansion,$\mathbit{P}\mathbit{V}\mathbf{-}\mathbit{n}\mathbit{R}\mathbit{T}\mathbf{\left(}\mathbf{1}\mathbf{+}\frac{\mathbf{B}\left(T\right)}{\left(V/n\right)}\mathbf{+}\frac{\mathbf{C}\left(T\right)}{{\left(V/n\right)}^{\mathbf{2}}}\mathbf{+}\mathbf{\cdots }\mathbf{\right)}$where the functions $\mathbit{B}\mathbf{\left(}\mathbf{T}\mathbf{\right)}$, $\mathbit{C}\mathbf{\left(}\mathbf{T}\mathbf{\right)}$, and so on are called the virial coefficients. When the density of the gas is fairly low, so that the volume per mole is large, each term in the series is much smaller than the one before. In many situations, it’s sufficient to omit the third term and concentrate on the second, whose coefficient $\mathbit{B}\mathbf{\left(}\mathbf{T}\mathbf{\right)}$ is called the second virial coefficient (the first coefficient is 1). Here are some measured values of the second virial coefficient for nitrogen (${\mathbit{N}}_{\mathbf{2}}$):$\mathbit{T}\mathbf{\left(}\mathbf{K}\mathbf{\right)}$$\mathbit{B}\mathbf{\left(}\mathbf{c}{\mathbf{m}}^{\mathbf{3}}\mathbf{/}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{\right)}$100–160200–35300–4.24009.050016.960021.3For each temperature in the table, compute the second term in the virial equation, $\mathbit{B}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{/}\mathbf{\left(}\mathbf{V}\mathbf{/}\mathbf{n}\mathbf{\right)}$, for nitrogen at atmospheric pressure. Discuss the validity of the ideal gas law under these conditions.Think about the forces between molecules, and explain why we might expect $\mathbit{B}\mathbf{\left(}\mathbf{T}\mathbf{\right)}$ to be negative at low temperatures but positive at high temperatures.Any proposed relation between $\mathbit{P}$, $\mathbit{V}$, and $\mathbit{T}$, like the ideal gas law or the virial equation, is called an equation of state. Another famous equation of state, which is qualitatively accurate even for dense fluids, is the van der Waals equation,$\mathbf{\left(}\mathbf{P}\mathbf{+}\frac{\mathbf{a}{\mathbf{n}}^{\mathbf{2}}}{{\mathbf{V}}^{\mathbf{2}}}\mathbf{\right)}\mathbf{\left(}\mathbf{V}\mathbf{-}\mathbf{n}\mathbf{b}\mathbf{\right)}\mathbf{=}\mathbit{n}\mathbit{R}\mathbit{T}$where a and b are constants that depend on the type of gas. Calculate the second and third virial coefficients ($\mathbit{B}$ and $\mathbit{C}$) for a gas obeying the van der Waals equation, in terms of $\mathbit{a}$ and $\mathbit{b}$. (Hint: The binomial expansion says that ${\mathbf{\left(}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{\right)}}^{\mathbf{p}}\mathbf{\approx }\mathbf{1}\mathbf{+}\mathbit{p}\mathbit{x}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbit{p}\mathbf{\left(}\mathbf{p}\mathbf{-}\mathbf{1}\mathbf{\right)}{\mathbit{x}}^{\mathbf{2}}$, provided that $\mathbf{|}\mathbf{p}\mathbf{x}\mathbf{|}\mathbf{\ll }\mathbf{1}$. Apply this approximation to the quantity ${\mathbf{\left[}\mathbf{1}\mathbf{-}\left(nb/V\right)\mathbf{\right]}}^{\mathbf{-}\mathbf{1}}$.)Plot a graph of the van der Waals prediction for $\mathbit{B}\mathbf{\left(}\mathbf{T}\mathbf{\right)}$, choosing $\mathbit{a}$ and $\mathbit{b}$ so as to approximately match the data given above for nitrogen. Discuss the accuracy of the van der Waals equation over this range of conditions. (The van der Waals equation is discussed much further in Section 5.3.)

Part (a). value of for different temperatures is shown in tabular form as shown below:

 $T\left(K\right)$ $B\left({\text{cm}}^{\text{3}}\text{/mol}\right)$ $B\left({\text{m}}^{\text{3}}\text{/mol}\right)$ $\frac{B\left(T\right)}{\left(V/n\right)}$ 100 -160 $-160×{10}^{-6}$ -0.0195 200 -35 $-35×{10}^{-6}$ -0.00213 300 -4.2 $-4.2×{10}^{-6}$ -0.00017 400 9.0 $9×{10}^{-6}$ 0.000274 500 16.9 $16.9×{10}^{-6}$ 0.000412 600 21.3 $21.3×{10}^{-6}$ 0.000433

Part (b). At high temperatures, molecular speed is high and the distance between the molecules increases. Whereas, at low temperature, molecular speed is low so the value of $B\left(T\right)$ is negative.

Part (c). Value of virial coefficient is $\overline{)B\left(T\right)=b-\frac{a}{RT}}$ and $\overline{)C\left(T\right)={b}^{2}}$.

Part (d). The value of a and b are $\overline{)a=0.1822}$ And $\overline{)b=6.42×{10}^{-5}}$ on the other hand, the fit is good.

See the step by step solution

## Part a. Step 1. Given.

Some measure value of the second virial coefficient for Nitrogen is given below:

 $T\left(K\right)$ $B\left({\text{cm}}^{\text{3}}\text{/mol}\right)$ 100 -160 200 -35 300 -4.2 400 9.0 500 16.9 600 21.3

## Part a. Step 2. Formula used.

The equation for an ideal gas is expressed as

$PV=nRT$

$\frac{V}{n}=\frac{RT}{P}$ …… (2)

Here, $P$ is the pressure of the gas, $V$ is the volume of the gas, $T$ is temperature given, $R$ is the gas constant and its value is $8.31{\text{\hspace{0.17em}J.mol}}^{\text{-1}}{\text{.K}}^{\text{-1}}$.

Assume that pressure is 1 atm i.e. $P=1\text{\hspace{0.17em}atm}=101325\text{\hspace{0.17em}Pa}$.

## Part a. Step 3. Calculation.

For $T=100\text{\hspace{0.17em}K}$

Substitute $8.31{\text{\hspace{0.17em}J.mol}}^{\text{-1}}{\text{.K}}^{\text{-1}}$ for $R$, $100\text{\hspace{0.17em}K}$ for $T$ and $101325\text{\hspace{0.17em}pascal}$ for $P$ in equation (2)

$\begin{array}{l}\frac{V}{n}=\frac{8.31×100}{101325}=0.0082{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}\\ \frac{V}{n}=0.0082{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}\end{array}$

Substitute $-160×{10}^{-6}$ for $B\left(T\right)$ and $0.0082{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}$ for $\frac{V}{n}$ to get the ratio $\frac{B\left(T\right)}{\left(V/n\right)}$

$\overline{)\frac{B\left(T\right)}{\left(V/n\right)}=\frac{-160×{10}^{-6}}{0.0082}=-0.0195}$

For $T=200\text{\hspace{0.17em}K}$

Substitute $8.31{\text{\hspace{0.17em}J.mol}}^{\text{-1}}{\text{.K}}^{\text{-1}}$ for $R$, $200\text{\hspace{0.17em}K}$ for $T$ and $101325\text{\hspace{0.17em}pascal}$ for $P$ in equation (2)

$\begin{array}{l}\frac{V}{n}=\frac{8.31×200}{101325}=0.0164{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}\\ \frac{V}{n}=0.0164{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}\end{array}$

Substitute $-35×{10}^{-6}$ for $B\left(T\right)$ and $0.0164{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}$ for $\frac{V}{n}$ to get the ratio $\frac{B\left(T\right)}{\left(V/n\right)}$

$\overline{)\frac{B\left(T\right)}{\left(V/n\right)}=\frac{-35×{10}^{-6}}{0.0164}=-0.00213}$

For $T=300\text{\hspace{0.17em}K}$

Substitute $8.31{\text{\hspace{0.17em}J.mol}}^{\text{-1}}{\text{.K}}^{\text{-1}}$ for $R$, $300\text{\hspace{0.17em}K}$ for $T$, and $101325\text{\hspace{0.17em}pascal}$ for $P$ in equation (2)

$\begin{array}{l}\frac{V}{n}=\frac{8.31×300}{101325}=0.0246{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}\\ \frac{V}{n}=0.0246{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}\end{array}$

Substitute $-4.2×{10}^{-6}$ for $B\left(T\right)$ and $0.0246{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}$ for $\frac{V}{n}$ to get the ratio $\frac{B\left(T\right)}{\left(V/n\right)}$

$\overline{)\frac{B\left(T\right)}{\left(V/n\right)}=\frac{-4.2×{10}^{-6}}{0.0246}=-0.00017}$

For, $T=400\text{\hspace{0.17em}K}$

Substitute $8.31{\text{\hspace{0.17em}J.mol}}^{\text{-1}}{\text{.K}}^{\text{-1}}$ for $R$, $400\text{\hspace{0.17em}K}$ for $T$ and $101325\text{\hspace{0.17em}pascal}$ for $P$ in equation (2)

$\begin{array}{l}\frac{V}{n}=\frac{8.31×400}{101325}=0.0328{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}\\ \frac{V}{n}=0.0328{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}\end{array}$

Substitute $9×{10}^{-6}$ for $B\left(T\right)$ and $0.0328{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}$ for $\frac{V}{n}$ to get the ratio $\frac{B\left(T\right)}{\left(V/n\right)}$

$\overline{)\frac{B\left(T\right)}{\left(V/n\right)}=\frac{9×{10}^{-6}}{0.0328}=-0.000274}$

For, $T=500\text{\hspace{0.17em}K}$

Substitute $8.31{\text{\hspace{0.17em}J.mol}}^{\text{-1}}{\text{.K}}^{\text{-1}}$ for $R$, $500\text{\hspace{0.17em}K}$ for $T$ and $101325\text{\hspace{0.17em}pascal}$ for $P$ in equation (2)

$\begin{array}{l}\frac{V}{n}=\frac{8.31×500}{101325}=0.0410{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}\\ \frac{V}{n}=0.0410{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}\end{array}$

Substitute $16.9×{10}^{-6}$ for $B\left(T\right)$ and $0.0410{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}$ for $\frac{V}{n}$ to get the ratio $\frac{B\left(T\right)}{\left(V/n\right)}$

$\overline{)\frac{B\left(T\right)}{\left(V/n\right)}=\frac{16.9×{10}^{-6}}{0.0410}=0.000412}$

For, $T=600\text{\hspace{0.17em}K}$

Substitute $8.31{\text{\hspace{0.17em}J.mol}}^{\text{-1}}{\text{.K}}^{\text{-1}}$ for $R$, $500\text{\hspace{0.17em}K}$ for $T$ and $101325\text{\hspace{0.17em}pascal}$ for $P$ in equation (2)

$\begin{array}{l}\frac{V}{n}=\frac{8.31×600}{101325}=0.0492{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}\\ \frac{V}{n}=0.0492{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}\end{array}$

Substitute $21.3×{10}^{-6}$ for $B\left(T\right)$ and $0.0492{\text{\hspace{0.17em}m}}^{\text{3}}\text{/mol}$ for $\frac{V}{n}$ to get the ratio $\frac{B\left(T\right)}{\left(V/n\right)}$

$\overline{)\frac{B\left(T\right)}{\left(V/n\right)}=\frac{21.3×{10}^{-6}}{0.0492}=0.000433}$

## Part a. Step 4. Conclusion.

Hence, all value of $\frac{B\left(T\right)}{\left(V/n\right)}$ for different temperatures is shown in tabular form as shown below:

 $T\left(K\right)$ $B\left({\text{cm}}^{\text{3}}\text{/mol}\right)$ $B\left({\text{m}}^{\text{3}}\text{/mol}\right)$ $\frac{B\left(T\right)}{\left(V/n\right)}$ 100 -160 $-160×{10}^{-6}$ -0.0195 200 -35 $-35×{10}^{-6}$ -0.00213 300 -4.2 $-4.2×{10}^{-6}$ -0.00017 400 9.0 $9×{10}^{-6}$ 0.000274 500 16.9 $16.9×{10}^{-6}$ 0.000412 600 21.3 $21.3×{10}^{-6}$ 0.000433

## Part b. Step 1. Introduction.

Some measured value of the second virial coefficient for nitrogen is given below

 $T\left(K\right)$ $B\left({\text{cm}}^{\text{3}}\text{/mol}\right)$ 100 -160 200 -35 300 -4.2 400 9.0 500 16.9 600 21.3

## Part b. Step 2. Explanation.

The gas molecules experience a weak attraction when they get close to each other. At low temperatures, the molecular speed is lower so that the attraction is felt more strongly. Thus, the molecules would tend to be close to each other even if the interaction doesn’t occur which results in a slightly smaller volume. From this fact, the negative value of $B\left(T\right)$ means a smaller volume.

At high temperatures, the molecular speed of molecules is high enough such that they will have a greater distance between the molecules. So, the value of $B\left(T\right)$ is positive in this case.

## Part b. Step 3. Conclusion.

At high temperatures, molecular speed is high and the distance between the molecules increases. Whereas, at low temperature, molecular speed is low so the value of $B\left(T\right)$ is negative.

## Part c. Step 1. Formula used

Van der Walls equation is:

$\left(P+\frac{a{n}^{2}}{{V}^{2}}\right)\left(V-nb\right)=nRT$ …… (1)

Here, $P$ is the pressure of the gas, $V$ is the volume of gas, $a\text{\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}b$ are constants that depend on the type of gas.

For a real gas, this equation is modified as

$PV=nRT\left[1+\frac{B\left(T\right)}{V/n}+\frac{C\left(T\right)}{{\left(V/n\right)}^{2}}+...\right]$ …… (2)

Here, $B\left(T\right)$ and $C\left(T\right)$ are virial coefficients and the first term is itself 1. When the density of the gas is low, the volume per mole is large; each term in the series is much smaller than the one before.

## Part c. Step 2. Calculation.

Rewrite equation (1) and simplified

$\left(P+\frac{a{n}^{2}}{{V}^{2}}\right)V\left(1-nb/V\right)=nRT$

$\begin{array}{l}\left(PV+\frac{a{n}^{2}}{V}\right)\left(1-\frac{nb}{V}\right)=nRT\\ \left(PV+\frac{a{n}^{2}}{V}\right)=nRT{\left(1-\frac{nb}{V}\right)}^{-1}\\ PV=nRT{\left(1-\frac{nb}{V}\right)}^{-1}-\frac{a{n}^{2}}{V}\\ PV=nRT{\left(1-\frac{nb}{V}\right)}^{-1}-\frac{a{n}^{2}}{V}×\frac{nRT}{nRT}\end{array}$

$PV=nRT\left[{\left(1-\frac{nb}{V}\right)}^{-1}-\frac{a{n}^{2}}{V}×\frac{1}{nRT}\right]$

$PV=nRT\left[\underset{part\text{\hspace{0.17em}}A}{\underset{⏟}{{\left(1-\frac{b}{\left(V/n\right)}\right)}^{-1}}}-\frac{a}{RT\left(V/n\right)}\right]$…… (3)

In the above equation (3) in part A, when it is expanded by assuming, $\frac{bn}{V}\ll 1$

${\left(1-\frac{b}{V/n}\right)}^{-1}=1+\frac{b}{V/n}+\frac{{b}^{2}}{{\left(V/n\right)}^{2}}$…… (4)

Substitute $1+\frac{b}{V/n}+\frac{{b}^{2}}{{\left(V/n\right)}^{2}}$ for ${\left(1-\frac{b}{V/n}\right)}^{-1}$ in equation (3)

$PV=nRT\left[1+\frac{b}{V/n}+\frac{{b}^{2}}{{\left(V/n\right)}^{2}}-\frac{a}{RT\left(V/n\right)}\right]$

$PV=nRT\left[1+\frac{1}{V/n}\left(b-\frac{a}{RT}\right)+\frac{{b}^{2}}{{\left(V/n\right)}^{2}}\right]$ …… (5)

Compare equation (5) and equation (2)

$\overline{)B\left(T\right)=b-\frac{a}{RT}}$

And

$\overline{)C\left(T\right)={b}^{2}}$

## Part c. Step 3. Conclusion.

Hence, the required value of the virial coefficient is $\overline{)B\left(T\right)=b-\frac{a}{RT}}$ and $\overline{)C\left(T\right)={b}^{2}}$.

## Part d. Step 1. Given.

Some measure value of the second virial coefficient for Nitrogen is given below

 $T\left(K\right)$ $B\left({\text{cm}}^{\text{3}}\text{/mol}\right)$ 100 -160 200 -35 300 -4.2 400 9.0 500 16.9 600 21.3

## Part d. Step 2. Formula used.

Van der Walls equation is:

$\left(P+\frac{a{n}^{2}}{{V}^{2}}\right)\left(V-nb\right)=nRT$ …… (1)

Here, $P$ is the pressure of the gas, $V$ is the volume of gas, $a\text{\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}b$ are constants that depend on the type of gas.

The expression for the viral coefficient is given

$\overline{)B\left(T\right)=b-\frac{a}{RT}}$ …… (2)

$\overline{)C\left(T\right)={b}^{2}}$ …… (3)

## Part d. Step 3. Calculation.

Fit the curve of equation (2) to data in the table above, The value of a and b from this fitting is

$\overline{)a=0.1822}$ And $\overline{)b=6.42×{10}^{-5}}$

The fit is fairly good, so the Val der Walls equation is a decent model for these data.

## Part d. Step 4. Conclusion.

Hence, the required value of a and b are $\overline{)a=0.1822}$ And $\overline{)b=6.42×{10}^{-5}}$ on the other hand, the fit is good.

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