Van der Waals and boiling point

In summary: It increases by a factor of 2.In summary, in this problem, a constant volume gas thermometer is used to measure the pressure of a gas at two different temperatures. The first temperature is the triple point of water and the second temperature is the boiling point of a liquid. After some gas is removed from the thermometer, the experiment is repeated and the pressures are measured again. By using the van der Waals equations of state, it is possible to solve for the boiling temperature of the liquid. The pressure does not double in the first experiment, but it does increase by a factor of 2.
  • #1
VoteSaxon
25
1

Homework Statement


The bulb of a constant volume gas thermometer is immersed in an ice/water/water vapour mixture at equilibrium and the recorded pressure is 0.400 atm. It is then immersed in a boiling liquid and the pressure is 0.844 atm. Sufficient gas is then removed from the bulb such that, when the experiment is repeated, the gas pressures are 0.200 atm and 0.401 atm respectively. Calculate the boiling temperature of the liquid.

Homework Equations


van der Waals equations of state are probably relevant here:
[p + a(n/V)2][V - nb] = nRT
∴ [p + (a/Vm2)][Vm - b] = RT

The Attempt at a Solution


I tried to arrange things better. If at first it was immersed in an "ice/water/water vapour mixture" then T = the triple point of water. So as I see it:

V = constant
T1 = 273.16 K, p1= 0.400 atm
T2 = boiling point, p2 = 0.844 atm
Some gas removed
T1 = 273.16 K, p1= 0.200 atm
T2 = boiling point, p2 = 0.401 atm

Observations:
  • From T1 to T2, the pressure of the gas doubles
  • After the gas is removed, the pressure is halved, despite there being the same conditions is before
I feel like the solution is in my head somewhere, but I've been rattling with it for eons without success. Any and all help very much appreciated.
 
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  • #2
Have you tried solving the van der Waals equation for P and then find an expression for P2 - P1?
 
  • #3
VoteSaxon said:

Homework Statement


[p + (a/Vm2)][Vm - b] = RT
You have four unknowns to solve for. You have 4 equations (2 for each experiment). Solve accordingly.
P.S. the pressure does not double in experiment 1.
 

1. What is the significance of Van der Waals forces in determining the boiling point of a substance?

Van der Waals forces, also known as London dispersion forces, are weak intermolecular forces that exist between all molecules. These forces play a crucial role in determining the boiling point of a substance, as they are responsible for the attraction between molecules. The stronger the Van der Waals forces, the higher the boiling point of a substance.

2. How does molecular size affect Van der Waals forces and boiling point?

The size of a molecule directly affects the strength of Van der Waals forces. Larger molecules have more electrons, which leads to a stronger temporary dipole moment and therefore, stronger Van der Waals forces. As a result, substances with larger molecules tend to have higher boiling points.

3. Can Van der Waals forces be broken?

Yes, Van der Waals forces are relatively weak and can be broken when enough energy is applied to the substance. This is why substances with weaker Van der Waals forces, such as gases, have lower boiling points compared to substances with stronger Van der Waals forces, such as liquids.

4. How does the shape of a molecule affect Van der Waals forces and boiling point?

The shape of a molecule also plays a role in the strength of Van der Waals forces. A molecule with a larger surface area will have more points of contact with other molecules, leading to stronger Van der Waals forces. This results in a higher boiling point compared to a molecule with a smaller surface area.

5. Can other factors besides Van der Waals forces affect the boiling point of a substance?

Yes, other factors such as intermolecular hydrogen bonding or the presence of polar or nonpolar groups can also affect the boiling point of a substance. These factors may contribute additional attractive forces between molecules, leading to a higher or lower boiling point depending on the specific interactions.

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