10.7: Non-ideal Gases
- Page ID
- 170552
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Theory
Exercise \(\PageIndex{1}\)
Non-ideal behavior for a gas is most likely to be observed under conditions of
- high temperature and low pressure.
- high temperature and high pressure.
- low temperature and low pressure.
- standard temperature and pressure.
- low temperature and high pressure.
- Answer
-
e. low temperature and high pressure.
Exercise \(\PageIndex{2}\)
Which of the following statements concerning real gases is/are CORRECT?
- Real gases are always liquids or solids at temperatures below 273.15 K.
- The pressure of a real gas is higher than predicted by the ideal gas law.
- The molecules in a real gas are attracted to each other.
a. 1 only b. 2 only c. 3 only d. 2 and 3 e. 1, 2, and 3
- Answer
-
c. 3 only
Exercise \(\PageIndex{3}\)
Real gases are those that
- only behave ideally at high pressures or low temperatures
- deviate from ideal behavior
- are only available naturally in the earth's atmosphere
- are called real gases because their behavior can easily be modeled
- have an even number of protons
- Answer
-
b. deviate from ideal behavior
Exercise \(\PageIndex{4}\)
Real gases deviate from ideal behavior because of intermolecular forces and _____.
- pressures within the chemical bonds
- nonzero molecular volumes
- ionization energies
- molecular vibrations
- a wide distribution of molecular speeds
- Answer
-
b. nonzero molecular volumes
Exercise \(\PageIndex{5}\)
The ideal gas law can be modified to correct for the errors arising from nonideality. The modified equation is known as the ________ equation of state for a real gas.
- Answer
-
van der Waals
van der Waals
Exercise \(\PageIndex{6}\)
The van der Waals constants for xenon are a = 4.19 (L2·atm)/mol2 and b = 0.0510 L/mol. If a 0.250 mol sample of xenon in a container with a volume of 3.65 L is cooled to −90°C, what is the pressure of the sample assuming ideal gas behavior? What would be the actual pressure under these conditions?
- Answer
-
Ideal Pressure: 1.03 atm; Actual Pressure: 1.01 atm
\(P = \frac{nRT}{V}\)
\(P = \frac{(0.250 mol)(0.08206 L*atm*mol^{-1}*K^{-1})(183.15 K)}{3.65 L}=1.029\;atm\)
\(P = \frac{nRT}{V-nb}-\frac{an^{2}}{V^{2}}\)
\(P = \frac{0.250*0.08206*183.15}{3.65-(0.250*0.0510)}-\frac{4.19*0.0250^{2}}{3.65^{2}}=1.0330-0.0196=1.0133\;atm\)

