Compressibility factor and van der Waals equation for temp

In summary: I guess you could obtain several different values of V for different temperatures with the help of a spreadsheet, then calculate each correspondig Z value and plot them against temperature. That'll give you some insight.
  • #1
sgstudent
739
3
The pressure exerted on the walls of the container by a real gas is less compared to an ideal gas. This is due to the attractive forces of the gas pulling the molecules back towards the rest of the gas molecules. However, there is also a relationship whereby at lower temperatures, the z is even lesser than at higher temperatures. I am trying to prove that using the van der Waals equation but I am having some trouble with that.

(P+a(n/v)^2 )(V-nb)=nRT

I wanted to compare 2 of the same gases at the same pressure but at different temperature would give us two different volumes. I tried to resolve V from the van der Waals equation but due to the cube on the equation I'm not sure how to evaluate the volume of the 2 gases at different temperatures.

PV/nRT = z = V/V-nb - (an/RTV)

Using the van der Waals equation, the above equation gives us the compressibility factor. I wanted to substitute the volume we obtain from the first equation and temperature to confirm that z is lower at lower temperatures. However because of the problem mentioned I am unable to do so.

Does anyone know how to use the van der Waals equation to explain why at lower temperatures the z is lower?
 
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  • #2
You could numerically solve the van der Waals cubic equation for V, and if it yields three real solutions you can choose the right one based on an approximate result gotten from a simpler EOS like the Ideal Gas Law. If it yields only one real solution, choosing the right result should be a no-brainer.
 
  • #3
MexChemE said:
You could numerically solve the van der Waals cubic equation for V, and if it yields three real solutions you can choose the right one based on an approximate result gotten from a simpler EOS like the Ideal Gas Law. If it yields only one real solution, choosing the right result should be a no-brainer.

Would it show that as temperature decreases the compressibility factor falls below 1 more than compared to when a higher temperature is used?
 
  • #4
sgstudent said:
Would it show that as temperature decreases the compressibility factor falls below 1 more than compared to when a higher temperature is used?
I guess you could obtain several different values of V for different temperatures with the help of a spreadsheet, then calculate each correspondig Z value and plot them against temperature. That'll give you some insight.
 

1. What is the compressibility factor?

The compressibility factor (Z) is a dimensionless quantity that describes how much a gas deviates from ideal gas behavior. It is the ratio of the actual volume of a gas to the volume it would occupy if it behaved as an ideal gas at the same temperature and pressure.

2. How is the compressibility factor related to temperature?

The compressibility factor is affected by temperature because it describes the deviation of a gas from ideal gas behavior. As temperature increases, the kinetic energy of gas molecules also increases, causing them to exert more pressure and occupy more volume. This results in a decrease in the compressibility factor.

3. What is the van der Waals equation for temperature?

The van der Waals equation is an equation of state that describes the behavior of real gases. It includes correction factors for intermolecular forces and the finite size of gas molecules. The equation is: (P + a/V^2)(V - b) = RT, where P is pressure, V is volume, T is temperature, a is a measure of intermolecular forces, and b is a measure of molecular size.

4. How does the van der Waals equation account for temperature?

The van der Waals equation takes into account the temperature of the gas through the ideal gas law, which is incorporated in the equation as RT. The ideal gas law states that the product of pressure and volume is directly proportional to the product of the number of moles and temperature. This allows the van der Waals equation to account for the effect of temperature on gas behavior.

5. Why is the van der Waals equation important for studying gases?

The van der Waals equation is important because it provides a more accurate description of the behavior of real gases compared to the ideal gas law. It takes into account factors such as intermolecular forces and molecular size, which are not considered in the ideal gas law. This allows for a better understanding and prediction of gas behavior, particularly at high pressures and low temperatures.

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