Is the compression factor of a gas equal to Vm/Vm-b?

Question

According to the definition, the compressibility factor is the ratio of the molar volume and the molar volume if the gas behaved as an ideal gas $$ Z = \frac{V_\mathrm{m}(p,T)}{V_\mathrm{m}^\pu{ig}(p,T)} \tag{1} $$

When comparing the ideal gas and the van der Waals equations of state, the ideal volume term is substituted by the real molar volume minus the correction factor $b$ accounting for the size of the molecules $$ p = \frac{RT}{\underbrace{V_\mathrm{m}^\pu{ig}}} \hspace{1 cm} p = \frac{RT}{\underbrace{V_\mathrm{m} - b}} - \frac{a}{V_\mathrm{m}^2} \tag{2} $$ So we can say that $$ V_\mathrm{m}^\pu{ig} = V_\mathrm{m} - b \tag{3} $$

Is this the case only when the constant $a \approx 0 $? Eq. (3) does not account for attraction forces, so it is obviously wrong. What is wrong with my logic?

Answer 1

To address the title question:

No, it is not, as the reality of real gases changes both volume and pressure. Therefore, if we corrected real volume to ideal volume, the real pressure would still differ from ideal pressure and vice versa.

It is approximately true for low pressure and not too low temperature, when repulsion due finite volume is (much) more significant than cohesive forces.

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$$a \ne 0 \implies V_\mathrm{m,Ideal} \ne V_\mathrm{m} - b$$

$$(p + \frac{a}{V_\mathrm{m}^2})(V_\mathrm{m}-b) = RT$$

$$(1 + \frac {a}{pV_\mathrm{m}^2})(1 -\frac{b}{V_\mathrm{m}}) = \frac 1Z$$

$$1 -\frac{b}{V_\mathrm{m}} + \frac {a}{pV_\mathrm{m}^2} - \frac {ab}{pV_\mathrm{m}^3}= \frac 1Z$$

As $1 + x \approx \frac {1}{1 - x}$ if $|x| \ll 1$:

$$Z \approx 1 +\frac{b}{V_\mathrm{m}} - \frac {a}{pV_\mathrm{m}^2} + \frac {ab}{pV_\mathrm{m}^3}\\=1 +\frac{b}{V_\mathrm{m}} - \frac {a}{pV_\mathrm{m}^2}(1 - \frac {b}{V_\mathrm{m}})$$

Answer 2

Real Volume

$$ \text{Real volume} \neq V- \mathrm{b} $$

You need to solve the van der Waals gas equation to obtain the real volume.

Compressibility Factor

According to your definition of compressibility factor $Z$:

$$ Z = \dfrac{V_\text{real}}{V_\text{ideal}} $$

and not the other way around; $Z \neq \left( \dfrac{V}{V-n\mathrm{b}} \right)$.

The van der Waals (Real) Gas Equation

$$ \left( P_\text{real}+\dfrac{\mathrm{a}n^2}{V_\text{real}^2} \right) \left( V_\text{real} - n\mathrm{b} \right) = n\mathrm{R}T \tag{1}$$

where $P_\text{real}$ and $V_\text{real}$ are the pressure and volume of the real gas, $n$ is the number of moles of gas particles, $\mathrm{a}$ and $\mathrm{b}$ are the van der Waals gas constants (which depend on the gas), and $\mathrm{R}$ is the universal gas constant.

Rearranging Equation (1) gives us:

$$ P_\text{real}\cdot V_\text{real}^3 - (n\mathrm{b}+nRT)\cdot V_\text{real}^2 + n^2\mathrm{a}\cdot V_\text{real}-n^3 \mathrm{ab} = 0 \tag{2} $$

Notice that this is a cubic equation and, thus, we will obtain three roots.

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Example

Here is an example of how $V_\text{real}$ varies with carbon dioxide gas $\ce{CO2}(g)$ for which $\mathrm{a} = \pu {3.658 bar L^2 mol^{-2}}$ and $\mathrm{b} = \pu {0.04286 L mol^{-1}}$. We are going to Assume:

$$ P_\text{real} = \pu{1 bar}\\ T = \pu{300 K}\\ n = \pu{1 mol} $$

Substituting in Equation (2):

$$ V_\text{real}^3 - 24.985\cdot V_\text{real}^2 + 3.658 \cdot V_\text{real}- 0.157 = 0 \\ \implies V_\text{real} = \pu{24.838 L} $$

The other two roots are complex and need not be considered. Under the same conditions:

$$ V_\text{ideal} = \dfrac{nRT}{P} = \pu{24.942 L}\\ \implies Z = \dfrac{V_\text{real}}{V_\text{ideal}} = \dfrac{24.838}{24.942} = 0.996 $$