Critical Pressure and Temperature of a van der Waals Gas

Waals equation of stateIn summary, the van der Waals equation of state can be used to calculate the critical temperature and pressure, which are given by T_{cr} = \frac{8a}{27bR} and P_{cr} = \frac{a}{27b^2}, respectively. To solve for these values, the P versus V curve must have an inflection point at the critical point, indicating that the first and second derivatives are equal to zero. This can be achieved by setting dP/dV = 0 and d^2P/dV^2 = 0, with the resulting equations being \frac{dP}{dV} = \frac{-RT}{n(V
  • #1
e(ho0n3
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Homework Statement


From the van der Waals equation of state, show that the critical temperature and pressure are given by

[tex]T_{cr} = \frac{8a}{27bR}[/tex]

[tex]P_{cr} = \frac{a}{27b^2}[/tex]

Hint: Use the fact that the [itex]P[/itex] versus [itex]V[/itex] curve has an inflection point at the critical point so that the first and second derivatives are zero.

Homework Equations


[tex]P = \frac{RT}{V/n - b} - \frac{a}{(V/n)^2}[/tex]

The Attempt at a Solution


The first and second derivative have powers of [itex]V[/itex] greater than 2. Unfortunately I don't have the skills to solve for [itex]dp/dt = 0[/itex] or [itex]d^2p/dt^2 = 0[/itex]. Perhaps there's a simpler way?
 
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  • #2
e(ho0n3 said:
The first and second derivative have powers of [itex]V[/itex] greater than 2. Unfortunately I don't have the skills to solve for [itex]dp/dt = 0[/itex] or [itex]d^2p/dt^2 = 0[/itex]. Perhaps there's a simpler way?

Err, that should be [itex]dP/dV = 0[/itex] and [itex]d^2P/dV^2 = 0[/itex].
 
  • #3
Just for reference,

[tex]\frac{dP}{dV} = \frac{-RT}{n(V/n - b)^2}[/tex]

[tex]\frac{d^2P}{dV^2} = \frac{2RT}{n^2(V/n - b)^3}[/tex]
 
  • #4
http://www.chem.arizona.edu/~salzmanr/480a/480ants/vdwcrit/vdwcrit.html

A good website with complete calculation
 

What is the critical pressure of a van der Waals gas?

The critical pressure of a van der Waals gas is the pressure at which the gas can no longer be liquefied, no matter how much the temperature is lowered. It is a characteristic property of a gas and is dependent on the intermolecular forces between the gas particles.

What is the critical temperature of a van der Waals gas?

The critical temperature of a van der Waals gas is the temperature at which the gas can no longer be liquefied, no matter how high the pressure is increased. It is a characteristic property of a gas and is dependent on the intermolecular forces between the gas particles.

How is the critical pressure and temperature of a van der Waals gas related?

The critical pressure and temperature of a van der Waals gas are related by the van der Waals equation, which is an improved version of the ideal gas law. The critical pressure and temperature can be calculated using the van der Waals constants, which are specific for each gas.

What happens to a van der Waals gas when it reaches its critical point?

When a van der Waals gas reaches its critical point, it undergoes a phase transition from gas to liquid. At this point, the gas and liquid phases become indistinguishable and the gas cannot be liquefied by changing the pressure or temperature. The critical point is also the point of maximum density for a van der Waals gas.

How does the critical pressure and temperature of a van der Waals gas compare to an ideal gas?

The critical pressure and temperature of a van der Waals gas are generally higher than that of an ideal gas. This is because the intermolecular forces in a van der Waals gas cause the gas particles to attract each other, making it more difficult for them to be compressed or condensed. This results in a higher critical pressure and temperature for van der Waals gases compared to ideal gases.

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