Van der Waals Equation Explained for Real Gases

The ideal gas law, represented as \(PV=nRT\), accurately describes gas behavior under conditions of low pressure and high temperature. However, real gases deviate from this ideal behavior due to two primary factors:

• Intermolecular Forces: Attractive forces exist between gas molecules, which reduce the force of collisions with the container walls, effectively lowering the observed pressure compared to the ideal prediction.
• Molecular Volume: Gas molecules themselves occupy a finite volume, reducing the space available for movement compared to the total volume of the container.

Understanding Van der Waals' Corrections

Johannes Diderik van der Waals introduced corrections to the ideal gas law to account for these real gas behaviors. His equation modifies both pressure ($p$) and volume ($V$) terms:

Pressure Correction Term

The term $\frac{a}{V_m^2}$ is added to the pressure term $p$. Here, $V_m$ represents the molar volume (volume per mole). This term corrects for the reduction in pressure caused by intermolecular attractive forces. The magnitude of this correction depends on the concentration of molecules (inversely proportional to molar volume squared) and the strength of attraction ($a$), a constant specific to each gas.

So, the pressure term becomes: $\left(p + \frac{a}{V_m^2}\right)$

Volume Correction Term

The term $b$ is subtracted from the molar volume $V_m$. This term represents the volume excluded by the molecules themselves, accounting for their finite size. It corrects for the fact that the available volume for gas molecules to move in is less than the total container volume.

So, the volume term becomes: $(V_m - b)$

The Complete Van der Waals Equation

Combining these corrected terms, the Van der Waals equation for one mole of a real gas is:

\(\rm \left(p + \frac{a}{V_m^2}\right)(V_m - b) = RT\)

Where:

• $p$ = Pressure of the gas
• $V_m$ = Molar volume of the gas
• $T$ = Absolute temperature
• $R$ = Ideal gas constant
• $a$ = Van der Waals constant related to intermolecular attractive forces
• $b$ = Van der Waals constant related to the volume excluded by molecules

Evaluating the Options

Let's examine the provided options against the established Van der Waals equation:

1. \(\rm \left(p+\frac{a}{v^2}\right)(v^2-b)=RT\): This option incorrectly modifies the volume term to $(v^2 - b)$.
2. \(\rm \left(p+\frac{a}{v^2}\right)(v-b)=RT^2\): This option incorrectly changes the temperature term to $RT^2$.
3. \(\rm \left(p+\frac{a}{v}\right)(v^2-b)=RT \): This option has incorrect pressure ($\frac{a}{v}$) and volume ($(v^2-b)$) correction terms.
4. \(\rm \left(p+\frac{a}{v^2}\right)(v-b)=RT\): This option correctly represents the Van der Waals equation, assuming '$v$' denotes the molar volume. The pressure is corrected by $\frac{a}{v^2}$ and the volume by $(v-b)$.
5. Option 5 is empty.

Therefore, the correct expression for Van der Waals' equation for a real gas among the choices is the fourth one.