Chapter 1, Problem 1.79E

Interpretation Introduction

(a)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Answer to Problem 1.79E

The molar volumes of (a) krypton, Kr is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for Krypton a = 2.318 atm L²/mol²;

b = 0.03978 L/mol

Boyle temperature, $\mathrm{T}\mathrm{b}=\mathrm{a}/\mathrm{b}\mathrm{R}$ = 711.04 K

Molar volume for krypton $\stackrel{}{\upsilon }=\mathrm{R}\mathrm{T}/\mathrm{p}$ = 59.1 L

Explanation of Solution

The non-ideal gas equation represented as;

$\left[p+a{n}^2/V^2\right]\left[V–nb\right]=nRT \dots \left(1\right)$

In the above equation,

[p+ an²/V²] Correction term introduced for molecular attraction

[V– nb] correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for Krypton a = $2.318atm{L}^2/mo{l}^2$

$b=0.03978L/mol$

Boyle temperature $\mathrm{T}\mathrm{b}\mathrm{ }=\mathrm{ }\mathrm{a}/\mathrm{b}\mathrm{R}$

= $\left(2.318atm{L}^{2}mo{l}^{-2}\right)/(0.03978Lmo{l}^{-1}x0.08205L.atm{K}^{-1}mo{l}^{-1}$

= 711.04 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of krypton the molar volume is, at one bar pressure

$\overline{\upsilon }\mathrm{ }=\mathrm{ }\mathrm{R}\mathrm{T}/\mathrm{p}$

$\begin{array}{l}=\left(0.08314Lbar{K}^{-1}mo{l}^{-1}x1molx711.04K\right)/1bar\hfill \\ =59.1L\hfill \end{array}$

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of krypton, Kr is calculated at 25 °C and 1 bar pressure.

Interpretation Introduction

(b)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of (b) ethane, C₂H₆ is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Answer to Problem 1.79E

The molar volumes of ethane, C₂H₆ is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for ethane a = 5.489 atm L²/mol²;

b = 0.0638 L/mol

Boyle temperature T_b = a/bR = 1049.5 K

Molar volume for ethane ῡ = RT/p = 87.2 L

Explanation of Solution

The non-ideal gas equation represented as;

$\left[p+a{n}^2/V^2\right]\left[V–nb\right]=nRT\dots \left(1\right)$

In the above equation,

[p + an²/V²] Correction term introduced for molecular attraction

[V – nb] Correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for ethane a = 5.489 atm L²/mol²

b = 0.0638 L/mol

Boyle temperature T_b = a/bR

$\begin{array}{l}=\left(5.489atm{L}^{2}mo{l}^{-2}\right)/(0.0638Lmo{l}^{-1}x0.08205L.atm{K}^{-1}mo{l}^{-1}\hfill \\ =1049.5K\hfill \end{array}$

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of ethane the molar volume is, at one bar pressure

ῡ = RT/p

$\begin{array}{l}=\left(0.08314Lbar{K}^{-1}mo{l}^{-1}x1molx1049.5K\right)/1bar\hfill \\ =87.2L\hfill \end{array}$

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of ethane, C₂H₆ is calculated at 25 °C and 1 bar pressure.

Interpretation Introduction

(c)

Interpretation:

Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is to be calculated at 25 °C and 1 bar pressure.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. It is generally denoted as the van der Waals equation of state.

Answer to Problem 1.79E

The molar volumes of mercury is calculated at 25 °C and 1 bar pressure as follows;

Van der Waals constant for mercury a = 8.093atm L²/mol²;

b = 0.01696 L/mol

Boyle temperature T_b = a/bR = 5822 K

Molar volume for mercury ῡ = RT/p = 484 L

Explanation of Solution

The non-ideal gas equation represented as;

$\left[p+a{n}^2/V^2\right]\left[V–nb\right]=nRT \dots \left(1\right)$

In the above equation,

[p + an²/V²] Correction term introduced for molecular attraction

[V – nb] Correction term introduced for volume of molecules

‘a’ and ‘b’ are called as van der Waals constants

a = the pressure correction and it is related to the magnitude and strength of the interactions between gas particles.

b = the volume correction and it is having relationship to the size of the gas particles.

Given;

Van der Waals constant for mercury a = 8.093atm L²/mol²;

b = 0.01696 L/mol

Boyle temperature T_b = a/bR

= (8.093 atm L² mol^-2)/(0.01696 L mol^-1 x 0.08205 L. atm K^-1 mol^-1

= 5822 K

At Boyle temperature, the second virial coefficient B is zero. Thus, for one mole of mercury the molar volume is, at one bar pressure

ῡ = RT/p

$\begin{array}{l}=\left(0.08314Lbar{K}^{-1}mo{l}^{-1}x1molx5822K\right)/1bar\hfill \\ =484L\hfill \end{array}$

Conclusion

Using the van der Waals constants given in Table 1.6, the molar volumes of mercury Hg is calculated at 25 °C and 1 bar pressure.