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The relationship between van der waals forces and hydrophobic interactions is that the van der waals act to bind the hydrophobe - non-polar substance - together, to separate from the polar solvent/water, and these contribute to the energy needed to separate the two substances. The seperation causes a decrease in the entropy of the system.
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The entropy change of a van der Waals gas expansion can be calculated using the formula ΔS = nRln (Vf/Vi) + ΔSint, where n is the number of moles of gas, R is the gas constant, Vf and Vi are the final and initial volumes of the gas, and ΔSint is the change in internal entropy due to the attractive forces between gas molecules.
It describes how the pressure, volume, and temperature of a gas are related. The heat capacities of a gas, on the other hand, refer to the amount of heat energy needed to raise the temperature of the gas. The van der Waals equation includes terms for the heat capacities of gases, allowing for a more accurate prediction of their behavior. 2.
The van der Waals equation isn't the best equation for corrections, particularly near the critical point for the gas. There are numerous other "real gas equations" which predict gas behavior better. (I'm not sure of what gas and what conditions, but there has to be a gas which has greater volume than would be predicted by ideal gas behavior.)
3) shape and size of a molecule are involved in the overall strength of local dispersive forces. This point influences also the B.P.,togheter with Mw. See London dispersion, Debye and Keesom forces, that together are covered by the broader term van der Waal's force, sometimes with Casimir effect. OTOH, the hydrogen bond is the true chemical bond.
Meanwhile steric strain (also known as van der Waals strain) can be thought as the repulsion when two bulky groups which are not directly bonded to each other become too close to each other and hence there isn't enough space for them. Here is the more detailed version. Torsional Strain. Lets consider an ethane molecule.
dU/dV in van der Waals equation of state derivation. 0. Enthalpy change for a van der waal gas. 7.
It may be based on purely elastic electrostatic interactions, where only the weakest van der Waals force may ( or even may not ) be involved. The probability 2 benzene molecules interact is bigger than for 2 water molecules. Notes also that for n-hexane is $\pu{a = 24.71 L^2\cdot bar mol^{2-}}$.
Can the van der Waals coefficients be negative in the van der Waals equation for real gases? Related. 20.