AP Chemistry : Gases
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Example Questions
Example Question #21 : Gases
Would you expect a polar or non polar gas to deviate most from ideal gas behavior?
Non polar gases, because of high dispersion interactions
Polar gases because of hydrogen bonding
Both polar and non polar gases behave ideally
Polar gases, because of high dipole-dipole interactions
Non polar gases because of reduced overall intermolecular forces
Polar gases, because of high dipole-dipole interactions
Polar gases would have increased interactions due to their dipoles that would lead to deviations from ideal gas behavior.
Example Question #22 : Gases
Which of the following would behave most like an ideal gas?
All are ideal gases because they are non-polar
Example Question #23 : Gases
Why do gases deviate from ideal behavior as the temperature is decreased?
As the temperature is decreased the molecules have less kinetic energy and can’t maintain the intermolecular forces necessary for ideal gas behavior.
As the temperatures is decreased the molecules become frozen in place.
As the temperature is decreased the molecules have more kinetic energy and break the intermolecular interactions keeping them together.
None of the above.
As the temperature is decreased the molecules have less kinetic energy and can’t break the intermolecular interactions between them.
As the temperature is decreased the molecules have less kinetic energy and can’t break the intermolecular interactions between them.
The ideal gas law assumes the gas particles are non-interacting and small relative to the size of their container. As the temperature is decreased the gas molecules are moving slower and allow for a greater degree of interaction.
Example Question #24 : Gases
When does a gas behave most like an ideal gas?
At low temperatures, low volume, low intermolecular interactions
At high temperature, high volumes, and high intermolecular interactions
At low volumes, high temperatures, and high intermolecular interactions
At high temperatures, high volume, low intermolecular interactions
At low temperatures, high volume, and low intermolecular interactions
At high temperatures, high volume, low intermolecular interactions
The ideal gas law assumes the gas particles are non-interacting and small relative to the size of their container. At high temperatures the gas molecules are moving fast enough to shorten the time scale for any interactions. At high volumes, the molecular size becomes small relative to the size of the container, and the low interactions mean the molecules act more independently.
Example Question #21 : Gases
What is the final pressure of a gas initially has a pressure of 10 atm at 50 L if the volume s now 25 L?
20 atm
10 atm
5 atm
25 atm
50 atm
20 atm
Use P1V1 = P2V2
P1 = 10atm; V1 = 50L
P2 = X; V2 = 25L
(10atm)(50L) = (x)(25L)
500 = 25x
x = 20
Example Question #1 : Ideal Gas Law
A sample of ideal gas is heated in a 2L vessel at a temperature of 320 Kelvin. The pressure in the vessel is 2.5atm. What is the new pressure in the vessel if the volume is halved and the temperature is reduced to 250 Kelvin?
Since the gas is ideal, we can use a variation of the ideal gas law in order to find the unknown final pressure.
Since we know that the number of moles is constant between both vessels (and R is a constant as well), we can simply compare the three factors being manipulated between the two vessels: pressure, volume, and temperature. Using a combination of Boyle's law and Charles's law, we can compare the two vessels to one another using the following equation.
Use the given values to solve for the final pressure.
Example Question #1 : Ideal Gas Law
An ideal gas takes up 60L at 2 atm. If the gas is compressed to 30L, what will the new pressure be?
Ideal gas law (modified)
P1V1 = P2V2
P1 = 2 atm; V1 = 60L; P2 = ?; V2 = 30L
(2)(60) = (X)(30)
P2 = 4 atm
Example Question #21 : Gases
Ammonia is created according to the balanced equation below.
The reaction is allowed to take place in a rigid container. Eight moles of hydrogen gas are mixed with two moles of nitrogen gas. The initial pressure exerted on the container is 5atm.
Assuming the reaction runs to completion, what will the pressure exerted on the vessel be after the reaction takes place?
Since the total pressure is dependent on the number of moles in the container, we can use the ratio of moles before and after the reaction to determine the final pressure in the container. There are initially ten moles of gas in the container, eight moles of hydrogen and two moles of nitrogen.
The next step is to determine how many moles of ammonia are created in the reaction, and if there is any excess reactant left over after the reaction. Since there is a 1:3 ratio for hydrogen gas to nitrogen gas, only six of the moles of hydrogen gas will be used in order to react will all two moles of the nitrogen gas.
This leaves two moles of excess hydrogen gas. Using stoichiometry and the molar ratios, we determine that four moles of ammonia are created in the reaction that comsumes two moles of nitrogen.
Four moles of ammonia, plus the two remaining moles of hydrogen gas, results in six moles of total gas after the reaction has run to completion.
Six moles is 60% of the inital moles in the container, so the final pressure will be 60% of the initial pressure. We can solve using the ideal gas law.
Example Question #3 : Ideal Gas Law
How many moles of Oxygen gas are present at a volume of 10 L at 1 atm and 25o C? (MW Oxygen gas = 32 g/mol)
0.041 mol
14 mol
0.41 mol
2.5 mol
4 mol
0.41 mol
use PV = nRT
n = PV/ RT
P = 1 atm; V = 10 L; R = 0.0821 Latm/molK; T = 298 K (MUST switch temperature to K)
n = moles of gas
n = (1atm)(10L)/(0.0821Latm/molK)(298K)
n= 0.41 mol
Example Question #4 : Ideal Gas Law
How many moles will a gas that is behaving ideally be if it takes up 2L at 4atm at 37o C?
0.314 mol
0.6 mol
2.63 mol
3.18 mol
0.379 mol
0.314 mol
Use PV = nRT
n = PV/RT
= (2L)(4atm) / (0.0821 Latm/molK)(310 K) <-- must change T into K
= 8/25.451
= 0.314 mol
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