Chapter 1, Problem 1.38E

Calculate the Boyle temperatures for carbon dioxide, oxygen, and nitrogen using the van der Waals constants in Table 1.6. How close do they come to the experimental values from Table 1.5?

Interpretation Introduction

Interpretation:

The Boyle temperatures for carbon dioxide, oxygen, and nitrogen using the van der Waals constants in Table 1.6. is to be calculated. And is to be compared to the experimental values from Table 1.5.

Concept introduction:

The Boyle temperature is defined as the temperature for which the second virial coefficient, becomes zero. In other words, the Boyle temperature is the temperature at which a non-ideal gas behaves like an ideal gas. In this temperature the attractive and repulsive forces acting on the gas molecules balance each out.

Answer to Problem 1.38E

The Boyle temperature of gas are as follows;

Entry	Gas	TB (K)(calculated values)	TB (K)(experimental values)	a (atm L2 / mol2)	B (L / mol)
1	Carbon dioxide	1026	713	3.592	0.04267
2	Oxygen	521	405	1.360	0.03183
3	Nitrogen	433	327	1.390	0.03913

The experimental values of Boyle temperature (T_B) are compared with calculated values and found to be in low.

Explanation of Solution

The Boyle temperature (T_B) is defined as the temperature for which the second virial coefficient, becomes zero. In other words, the Boyle temperature is the temperature at which a non-ideal gas behaves like an ideal gas. In this temperature the attractive and repulsive forces acting on the gas molecules balance each out. Thus ‘T_B^’ is given by the expression

${\text{T}}_B=\frac{a}{bR}\cdot \cdot \cdot (1)$

‘a’ and ‘b’ are called as van der Waals constants and ‘a’ represents the pressure correction and it is related to the magnitude and strength of the interactions between gas particles. Similarly, ‘b’ describes the volume correction and it is having relationship to the size of the gas particles.

$p=RT\left(\frac{1}{\overline{V}}+\frac{B_2(T)}{{\overline{V}}^2}+\frac{B_3(T)}{{\overline{V}}^3}+\cdot \cdot \cdot \right)\cdot \cdot \cdot (2)$

Since, the Boyle temperature is the temperature for which second virial coefficient becomes 0. Moreover, it is at this temperature that the forces of attraction and repulsion acting on the gas molecules balance out. Higher order be virial coefficients are smaller than the second virial coefficient, the gas likes to behave as an ideal gas over a entire range of pressures. At low pressures,

the equation transforms as,

$\frac{dZ}{dP}=0\text{If}p=0$

Where, Z = Compressibility factor.

T_B of carbon dioxide

Knowing a and b values of CO₂ from table 1.6 T_B of CO₂ is calculated as follows;

$a = 3.592 \mathrm{a}\mathrm{t}\mathrm{m}. \mathrm{L}2 / \mathrm{m}\mathrm{o}\mathrm{l}2$

$b = 0.04267 \mathrm{L}\mathrm{ }/\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}$

R = 0.0823 L. atm / K. mol

$\begin{array}{l}{T}_B=\frac{3.592\text{atm}{\text{.L}}^{\text{2}}{\text{/mol}}^{\text{2}}}{(0.0823\text{L}\text{.atm/K}\text{.mol})(0.04267\text{L/mol})}\\ T_B=\frac{3.592}{823\times {10}^{-4}\times 4267\times {10}^{-5}}\\ T_B=1026\text{K}\end{array}$

T_B of oxygen:

Knowing a and b values of oxygen from table 1.6 T_B of oxygen is calculated as follows;

$a = 1.360 \mathrm{a}\mathrm{t}\mathrm{m}.\mathrm{ }\mathrm{L}2\mathrm{ }/\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}2$

$b = 0.03183 \mathrm{L}\mathrm{ }/\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}$

$R = 0.0823 \mathrm{L}.\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{m}\mathrm{ }/\mathrm{ }\mathrm{K}.\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}$

$\begin{array}{l}{T}_B=\frac{1.360\text{atm}{\text{.L}}^{\text{2}}{\text{/mol}}^{\text{2}}}{(0.0823\text{L}\text{.atm/K}\text{.mol})(0.03183\text{L/mol})}\\ T_B=\frac{1.360}{823\times {10}^{-4}\times 4267\times {10}^{-5}}\\ T_B=521\text{K}\end{array}$

T_B of nitrogen:

Knowing a and b values of nitrogen from table 1.6 T_B of nitrogen is calculated as follows;

$a = 1.390 \mathrm{a}\mathrm{t}\mathrm{m}.\mathrm{ }\mathrm{L}2\mathrm{ }/\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}2$

$b = 0.03913 \mathrm{L}\mathrm{ }/\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}$

$R = 0.0823 \mathrm{L}.\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{m}\mathrm{ }/\mathrm{ }\mathrm{K}.\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}$

$\begin{array}{l}{T}_B=\frac{1.390\text{atm}{\text{.L}}^{\text{2}}{\text{/mol}}^{\text{2}}}{(0.0823\text{L}\text{.atm/K}\text{.mol})(0.031913\text{L/mol})}\\ T_B=\frac{1.390}{823\times {10}^{-4}\times 3913\times {10}^{-5}}\\ T_B=433\text{K}\end{array}$

Moreover, the experimental Boyle temperature (table 1.5) values are compared with calculated values and the experimental values are found to be low.

Conclusion

Thus, the Boyle temperatures for carbon dioxide, oxygen, and nitrogen using the van der Waals constants in Table 1.6. is calculated. And compared to the experimental values from Table 1.5.