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Real Gases and the Van der Waals Equation

Ideal Gases


Wize Tip
Ideal gases have negligible volume.

And there are no intermolecular forces between gas particles.

Note: PV=nRT is the IDEAL gas law and describes IDEAL gases

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Real Gases

Wize Tip
Real gases have volume.

And they have intermolecular forces (attraction between gas molecules).

Real gases are most like ideal gases when temperature is high (well above the condensation temp for that gas), pressure is low (usually less than 1 atm)
Photo by SMART Servier Medical Art / CC BY

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Wize Concept
Real gases deviate more from ideal gases when temperature is low (near bp) and pressure is high (~>10atm)

  • With a lower temperature and higher pressure just think that we are going more towards liquids than gases
  • High pressures help to compress and push molecules closer together, resulting in less distance between molecules. As a result, the volume of the gas particles becomes more significant.
  • Both of these factors (low temp and high pressure) cause gas molecules to interact more with each other (have intermolecular forces) and interact less with the walls of the container (decreasing pressure).

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Ideal vs Real Gas in a Constant Pressure Container (Volume of the Container Can Change)

Ideal Gas

  • Has no intermolecular forces between gas particles
  • Has no volume
  • Container remains unchanged

Real Gas

  • Real gases have a significant volume!
  • Since they have volume, real gas particles want to interact with other gas particles (intermolecular attractions)
  • This interaction causes the particles to come closer together and the volume of the container decreases

Wize Concept
For real gases:
V(gas) > 0
V(container) less than the volume of a container for an ideal gas.


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Ideal vs Real Gas in a Constant Volume Container (Pressure in the Container Can Change)

Ideal Gas

  • Ideal gases have no intermolecular forces

Real Gas

  • Real gases have intermolecular forces
  • The intermolecular forces between real gas particles causes real gas particles to spend (more/less)
    more
    time close to other gas particles compared to ideal gases
  • As a result, real gas particles hit the walls of the container (more/less)
    less
    and with (more/less)
    less
    force
  • The pressure inside the container for a real gas is therefore (higher/lower)
    lower
    than the pressure inside the container for an ideal gas!

Wize Concept
Real gases have intermolecular attractions and this results in the pressure of the container decreasing.


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Van der Waals Equation


ideal gasesPV=nRTideal \ gases \rightarrow PV=nRT

real gases(P+an2V2)(Vnb)=nRTreal \ gases \rightarrow \boxed{\Big(P+\orange a\frac{n^2}{V^2} \Big)(V-n\orange b)=nRT}


For real gases, we need to adjust the ideal gas law. The adjusted version is called the Van der Waals equation.

Real gas particles are attracted to each other (they have intermolecular attractions).

  • In the Van der Waals equation, to "adjust" the equation to be more like ideal gases there is a constant, "a"
  • a=attractive force correction
  • a increases with increasing intermolecular attraction between molecules

Real gas particles also have volume.

  • In the Van der Waals equation, to "adjust" the equation to be more like ideal gases there is a constant, "b"
  • b=bulkiness correction
  • b increases with molecular weight of gas molecule (more volume)

Note: a and b are Van der Waals constants and experimentally determined parameters that are different for every gas.


Wize Tip
Know the differences between ideal and real gases (you are most likely to be tested on this)
Know what the constants a and b mean in the Van der Waals equation.
You don't need to memorize the Van der Waals equation :)

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Real Gases vs Ideal Gases & The Compressibility Factor

Review: There are two main differences between real gases and ideal gases: (**you should know this!!)

  • Real gases have volume
  • Real gases have intermolecular forces (particles are attracted to each other)

Review: Real gases are more likely to deviate from ideal behaviour under what conditions? (**you should know this!!)

At
high
pressure and
low
temperature real gases deviate most from ideal gas behaviour.

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Let's interpret the graph based on our knowledge!
  • PV=nRT
  • PV/RT=n so the y axis is n and on the graph we have ideal gas labelled as 1 mol
  • In lower temperatures did we say we would see more/less of a deviation from an ideal gas?
    More!
  • This is why we see the 200K line curve farthest away from the horizontal line representing an ideal gas!

Compressibility Factor (Z)

  • We can calculate how close a real gas behaves like an ideal gas by measuring the compressibility factor, Z.
  • For an ideal gas, Z=1
  • In this equation V is the molar volume (V/n)
Z=PVRT\boxed{Z=\frac{PV}{RT}}


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Example: Van der Waals Equation

3 moles of He gas are released into a 4 L container at 275 K. If He is a real gas, what is the pressure of the gas? (For He, a = 0.035 L2atmmol2\frac{L^{2}atm}{mol^{2}}, and b = 0.2371 L/mol). * Please note the mistake in the video for the units of a



As this is a real gas, we need to setup the Van der Waal's equation:

(P +  n2aV2)(Vnb) = nRT\left(P\ +\ \frac{\ n^2a}{V^2}\right)\cdot\left(V-nb\right)\ =\ nRT


Next, we will isolate for P:

(P +  n2aV2) = nRTVnb\left(P\ +\ \frac{\ n^2a}{V^2}\right)\ =\ \frac{nRT}{V-nb}

P= nRTVnb   n2aV2P=\ \frac{nRT}{V-nb}\ -\ \frac{\ n^2a}{V^2}

And now we fill in our terms:

P= (3)(0.08206 )(275)(4)[(3)(0.2371)]   (3)2(0.035)(4)2P=\ \frac{\left(3\right)\left(0.08206\ \right)\left(275\right)}{\left(4\right)-\left[\left(3\right)\cdot\left(0.2371\right)\right]}\ -\ \frac{\ \left(3\right)^2\cdot\left(0.035\right)}{\left(4\right)^2}

P =67.69953.2887   0.0196875P\ =\frac{67.6995}{3.2887}\ \ -\ 0.0196875

P = 20.58548971  0.0196875P\ =\ 20.58548971\ -\ 0.0196875

P = 20.57 atmP\ =\ 20.57\ atm


Example: Real vs Ideal Gases

Find the difference between the ideal pressure and the Van der Waals pressure for 2 mol of CO2 in a 1L vessel at room temperature. CO2 (a = 3.640 L2atm/mol2; b = 0.04267 L/mol).



1) Solve for pressure using the ideal gas law



P=nRTV=(2 mol)(0.08206 L atm K1mol1(298 K)1L=48.9 atmP=\frac{nRT}{V}=\frac{(2\ mol)(0.08206\ L\ atm\ K^{-1}mol^{-1}(298\ K)}{1L}=48.9\ atm


2) Solve for the Van der Waals pressure using the Van der Waals equation for real gases



P=nRT(Vnb)an2V2P=\frac{nRT}{(V-nb)}-a\frac{n^2}{V^2}

P=(2 mol)(0.08206 L atm K1mol1)(298 K)(1 L(2 mol)(0.04267Lmol)(3.640 atm L2 mol2)(2 mol)2(1 L)2P=\frac{(2\ mol)(0.08206\ L\ atm\ K^{-1}mol^{-1})(298\ K)}{\Big(1\ L-(2\ mol)(0.04267\frac{L}{mol}\Big)}-(3.640\ atm\ L^2\ mol^{-2})\frac{(2\ mol)^2}{(1\ L)^2}

P=38.9 atmP=38.9\ atm
We can see that at high pressures large deviations can occur.

3) Find the difference between the two pressures we solved for:


Difference in pressure=48.9atm - 38.9atm
Difference in pressure=10atm

Out of the following, which gas would deviate the most from ideal behavior at a low temperature and high pressure?



Extra Practice