There are two main assumptions for an ideal gas (and a few smaller assumptions). First, the gas particles of the ideal gas must have no molecular volume. Second, the gas particles must exert no intermolecular forces on each other; therefore, forces such hydrogen bonding, dipole-dipole interactions, and London dispersion forces are irrelevant in ideal gases. Other small assumptions of ideal gases include random particle motion (no currents), lack of intermolecular interaction with the container walls, and completely elastic collisions (a corollary of zero intermolecular forces).

For real gases, however, these assumptions are invalid. This means that the real gas particles have molecular volume and exert intermolecular forces on each other. 

Recall that the volume in the ideal gas law is the volume of the free space available inside the container. For ideal gases, the free space volume is equal to the volume of the container because the gas particles take up no volume; however, for real gases, the free space volume is the volume of the container minus the volume of the gas particles. Though the exact values of free space volume will differ, the volume of the container is important for both real and ideal gases.

Henry's Law directly applies to gas exchange in the lungs, as it states that the amount of dissolved gas in a liquid are proportional to the solubility of the gas in the liquid. As oxygen is dissolved in the bloodstream, it is able to diffuse into the air of the lungs depending upon intrapleural pressure of that particular gas. Essentially, Henry's law dictates how readily oxygen will cross the alveolar epithelium.

To solve this question you need to know the Dalton’s Law of partial pressure:

In this equation, is the partial pressure of the gas,  is the mole fraction of the gas, and  is the total pressure of the system. You also need to know the definition of mole fraction: moles of a compound divided by total moles in the system.

This question requires us to calculate the partial pressure of both gas B and gas C. First, let’s find the partial pressure of gas B. The mole fraction of gas B is:

We are given the total pressure of the system. Using the calculated mole fraction, we can solve for the partial pressure of gas B:

Similarly, we can find the partial pressure of gas C. The mole fraction of gas C is:

The mole fraction times the total pressure will give the partial pressure of gas C:

Finally, find the difference between the partial pressures of the two gases.

The partial pressure of gas B is  less than the partial pressure of gas C.

Recall Dalton’s Law of partial pressure:

In this equation,  is the partial pressure of the gas,  is the mole fraction of the gas, and  is the total pressure of the system. The mole fraction, , is defined as the moles of a compound divided by the total moles in the system:

From the given information, we can solve for mole fraction of gas A using the partial pressure of gas A and the total pressure of the system.

To find the mass of gas A we need to find the moles of gas A in the mixture and multiply it by the molecular weight of gas A; however, the information given in this question only allows us to solve for the mole fraction of gas A, or percentage of gas A in the system. To convert mole fraction to moles we would need to know the total number of moles in the mixture; therefore, the mass of gas A cannot be determined.

Partial pressure is given by the mole fraction of a gas multiplied by the total pressure in the container. Mole fraction is equal to the moles of a given compound divided by the total moles.

This is an application of Dalton's Law.

Partial pressure of a gas is defined as the pressure of the gas in a mixture of gases. It is calculated by using Dalton’s Law of partial pressure, which states that the partial pressure is equal to the product of the mole fraction of the gas and the total pressure of the system.

One of the assumptions of this law is that the gases are independent and do not chemically react with each other. This means that the gases in the system have to be inert. If the gases were reactive with each other, then Dalton’s Law would become invalid. Recall that Earth’s atmosphere contains about 78% nitrogen, 21% oxygen, and 1% all other gases; therefore, nitrogen has the largest partial pressure in the atmosphere because it has the greatest mole fraction.

Partial pressure does signify the pressure of a gas in a mixture of gases, but it is only valid for the temperature of the system. The partial pressure of the gas will be different for different temperatures. This occurs because pressure is dependent on temperature. Changing the temperature will not alter the mole fraction (amount of gas); however, it will alter the total pressure which will subsequently alter the partial pressure of the gas according to Dalton's Law.

The best answer is the compound's boiling point. At this point, it ceases obeying the gas laws and becomes a liquid.

If it were possible to remain in a gaseous state to zero Kelvin (), then this would be the correct answer. As the gas changes to a liquid, it becomes effectively resistant to further compression and will occupy the same volume through any further decrease in temperature.

This is also a straightforward question that can be answered from little to no information from the passage. The point of emphasis here is equilibrium. At a diffusion equilibrium there is no net change, and thus, a one-to-one oxygen molecule exchange fits the bill. 

To solve this question, we can use Graham's law of effusion:

Graham's law demonstrates that the rate of effusion is inversely proportional to the square root of the molar mass. Essentially, lighter gases will effuse more quickly than heavier gases.

Given the densities and equal volumes of the two gases, we see that the gas with the greater density will account for more molar mass. The volume of gas Y weighs twice as much as the volume of gas X. Based on this information, we can conclude that gas X diffuses faster because it has a lower molar mass.

Boyle's law describes the inverse relationship between a change and pressure and a change in volume, while a sample of gas is kept at constant temperature. This relationship can be mathematically written as:

or