Chemistry, asked by musain21711, 1 month ago

Derive the van der Waals' equation for a gas.Explain why van der Waals'
equation cannot be considered as a
gas.

generalized equation of state for real gases​

Answers

Answered by ramdasianiket8888
0

Answer:

In chemistry and thermodynamics, the Van der Waals equation (or Van der Waals equation of state; named after Dutch physicist Johannes Diderik van der Waals) is an equation of state that generalizes the ideal gas law based on plausible reasons that real gases do not act ideally. The ideal gas law treats gas molecules as point particles that interact with their containers but not each other, meaning they neither take up space nor change kinetic energy during collisions (i.e. all collisions are perfectly elastic).[1] The ideal gas law states that volume (V) occupied by n moles of any gas has a pressure (P) at temperature (T) in kelvins given by the following relationship, where R is the gas constant:

{\displaystyle PV=nRT}PV=nRT

To account for the volume that a real gas molecule takes up, the Van der Waals equation replaces V in the ideal gas law with {\displaystyle (V_{m}-b)}{\displaystyle (V_{m}-b)}, where Vm is the molar volume of the gas and b is the volume that is occupied by one mole of the molecules. This leads to:[1]

{\displaystyle P(V_{m}-b)=RT}{\displaystyle P(V_{m}-b)=RT}

Explanation:

The second modification made to the ideal gas law accounts for the fact that gas molecules do in fact interact with each other (they usually experience attraction at low pressures and repulsion at high pressures) and that real gases therefore show different compressibility than ideal gases. Van der Waals provided for intermolecular interaction by adding to the observed pressure P in the equation of state a term {\displaystyle a/V_{m}^{2}}{\displaystyle a/V_{m}^{2}}, where a is a constant whose value depends on the gas. The Van der Waals equation is therefore written as:[1]

{\displaystyle \left(P+a{\frac {1}{V_{m}^{2}}}\right)(V_{m}-b)=RT}{\displaystyle \left(P+a{\frac {1}{V_{m}^{2}}}\right)(V_{m}-b)=RT}

and can also be written as the equation below:

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

where Vm is the molar volume of the gas, R is the universal gas constant, T is temperature, P is pressure, and V is volume. When the molar volume Vm is large, b becomes negligible in comparison with Vm, a/Vm2 becomes negligible with respect to P, and the Van der Waals equation reduces to the ideal gas law, PVm=RT.[1]

It is available via its traditional derivation (a mechanical equation of state), or via a derivation based in statistical thermodynamics, the latter of which provides the partition function of the system and allows thermodynamic functions to be specified. It successfully approximates the behavior of real fluids above their critical temperatures and is qualitatively reasonable for their liquid and low-pressure gaseous states at low temperatures. However, near the phase transitions between gas and liquid, in the range of p, V, and T where the liquid phase and the gas phase are in equilibrium, the Van der Waals equation fails to accurately model observed experimental behaviour, in particular that p is a constant function of V at given temperatures. As such, the Van der Waals model is not useful only for calculations intended to predict real behavior in regions near the critical point. Corrections to address these predictive deficiencies have since been made, such as the equal area rule or the principle of corresponding states.

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