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Test: Gas Phase - MCAT MCQ


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10 Questions MCQ Test - Test: Gas Phase

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Test: Gas Phase - Question 1
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How many diagrams showing relationships between pressure, volume, and temperature below is/are incorrect for ideal gases?

Detailed Solution for Test: Gas Phase - Question 1
  • Although the graphs may look daunting, all of the relationships can be derived using the ideal gas law, PV=nRT.
  • Just rearrange the variables in the gas law so we can see if there is a positive correlation (on different sides of the equal sign) or a negative correlation (on the same side).
  • For A: P⋅V/T = nR, so this will be a negative correlation. A is false. For B: P = nR⋅T/V. B is true. For C: V = nR⋅T/P. C is true. For D: PV = nR⋅T. D is true.
  • Only one diagram shows an incorrect relationship for ideal gases.
Test: Gas Phase - Question 2
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At a given temperature T and pressure P, a person's lung holds a set volume of V of air. Given that air is roughly 20% oxygen, how many moles of oxygen are in his lungs?

Detailed Solution for Test: Gas Phase - Question 2
  • Use the ideal gas law to find n (moles of air).
  • n = PV/RT. This includes a mixture of nitrogen, oxygen, and other gases. From the mole fraction rule, we can infer that if the air is 20% oxygen, 20% of the total moles must be oxygen molecules.
  • Thus, there are 0.2*(PV/RT) moles of oxygen in his lungs.
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Test: Gas Phase - Question 3
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Gases found in the environment are most likely to exhibit properties similar to that of ideal gases under conditions of:

Detailed Solution for Test: Gas Phase - Question 3
  • Think back to the assumptions of the ideal gas laws.
  • In order to obey the ideal gas laws, we would need to minimize the effects of intermolecular interactions and molecular size of these gases. Thus, the ideal conditions will be under high temperatures (lots of kinetic energy in the molecules to move around and not be experiencing forces from each other) and low pressure (less inhibited to move around).
  • Gases found in the environment are most likely to exhibit properties similar to that of ideal gases under conditions of high temperatures and low pressures.
Test: Gas Phase - Question 4
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Assume an experiment vessel which is well fitted with a piston. The airtight piston has negligible mass and can move up and down freely. There is 1 mol of an ideal gas contained within the vessel under the piston at pressure of 100 kPa. The piston rests at 10 cm from the base of the vessel. The pressure in the vessel increases to 200 kPa as force is applied to the piston. What would be the new resting position of the piston from the base of the vessel? Assume the temperature is held constant at 300 K throughout the experiment.
Detailed Solution for Test: Gas Phase - Question 4
  • Try using Boyle's Law for this question.
  • P1​V1 ​= P2​V2​ , then V2 = P1V1/P2. It's important to realize that to find the volume occupied by the gas in this experiment vessel, we would use A (area of the piston) multiplied by h (the height of the piston). We then would have Ah2​ = P1​Ah1​/P2​. The surface areas actually cancel out, and you will have h2 = P1h1/P2.
  • Substituting in the values, you will get the new resting position as 5 cm.
Test: Gas Phase - Question 5
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Which properties reflected in real gases does the van der Waals equation attempt to account for by modifying the ideal gas law?
I. Volume
II. Pressure
III. Temperature
Detailed Solution for Test: Gas Phase - Question 5
  • Real gases experience intermolecular forces between gas molecules (van der Waals forces).
  • Real gases molecules possess mass.
  • Given that real gases experience intermolecular forces and possess mass, the van der Waals equation attempts to modify the ideal gas law by introducing the constants a and b, which adjusts for the effects of volume and pressure in real gases. (Real gases have lower pressure and higher volume compared to ideal gases) There is no term in the van der Waals equation that adjusts for temperature.
Test: Gas Phase - Question 6
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The behavior of which of these real gases will be reflected most closely with the ideal gas law?
Detailed Solution for Test: Gas Phase - Question 6
  • Consider the factors that distinguish real gases from ideal gases (i.e. what the van der Waals equation attempts to take into account)
  • Large gas molecules generally possess more volume and intermolecular forces.
  • Carbon dioxide and methane are large molecules, which immediately means more mass in each molecule, causing deviations from ideal gas behavior. Between helium and argon (both monomolecular noble gases), helium is the smaller gas molecule of the two, which similarly will have less of a mass effect, demonstrating closer adherence to the ideal gas law.
Test: Gas Phase - Question 7
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Assuming ideal gas properties, which of the following occupies the most volume at 273 K and 1 atm of pressure?
Detailed Solution for Test: Gas Phase - Question 7
  • 273 K and 1 atm of pressure is known as STP, or standard temperature and pressure.
  • At STP, all gases occupy 22.4L volume regardless of the identity of the gas.
  • One mole of nitrogen gas, oxygen gas and hydrogen gas all occupy the same volume at STP assuming ideal gas properties.
Test: Gas Phase - Question 8
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Atmospheric air is comprised, roughly, of 80% nitrogen and 20% oxygen. A 100 L sample of atmospheric air is kept at 300 K and 100 kPa. How many moles of oxygen molecules are found in this gas sample? (Use R = 10)
(L⋅kPa)/(mol⋅K))
Detailed Solution for Test: Gas Phase - Question 8
  • The law of partial pressures states that the volume of each gas in the mixture is proportional to the partial pressures of each gas in the mixture.
  • Given atmospheric air is 20% oxygen, the partial pressure of oxygen in this sample is 20 kPa. Now use the ideal gas equation to find n.
  • n = PV/RT,orn = 100⋅20/10⋅300 = 2/3moles.
Test: Gas Phase - Question 9
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Shown below are four mercury barometers of the same height (all four barometer tubes measure one meter from the tube opening to rounded top). Which barometer shows the greatest external pressure?
Detailed Solution for Test: Gas Phase - Question 9
  • Consider the difference in shapes between the different barometers.
  • Given that the force acting on the fluid in the barometer is F = mg. We can derive it thus: m = density⋅volume, so F = density⋅volume⋅g. Volume = area⋅height, so F=density⋅area⋅height⋅g. We know that F/area = pressure, so we can say that the change in pressure = the change in height⋅density⋅g.
Test: Gas Phase - Question 10
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All of the following are properties of ideal gases except:
Detailed Solution for Test: Gas Phase - Question 10
  • The property listed in option 1, "Small amounts of energy are lost during collisions between gas molecules," is not a characteristic of ideal gases. In reality, energy can be lost or gained during collisions between gas molecules. Ideal gases are a theoretical concept that assumes certain simplifying assumptions to make calculations and predictions easier.
  • One of these assumptions is that gas molecules collide elastically, as stated in option 3. This means that during collisions, there is no loss of kinetic energy, and the total energy of the gas remains constant. However, in real gases, some energy can be transferred in the form of heat or converted into other forms, leading to a loss of energy.
  • In an ideal gas, the molecules are considered to have negligible volume compared to the total volume occupied by the gas, as mentioned in option 2. This assumption allows us to ignore the individual volumes of gas molecules and treat the gas as a collection of point particles.
  • Option 4 states that gas molecules do not interact with each other except during collisions. This assumption implies that there are no attractive or repulsive forces between gas molecules. In reality, intermolecular forces can exist between gas molecules, although they may be weak in some cases.
  • In summary, option 1 is not a property of ideal gases because ideal gases assume completely elastic collisions, while in reality, some energy can be lost or gained during collisions between gas molecules.
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