This question asks you how the velocity of the molecules changes with increasing temperature. From molecular kinteics, we know that temperature is proportional to kinetic energy.

\(\displaystyle Kinetic\ Energy=\frac{3}{2}KT\)

We can see that as temperature rises, the kinetic energy rises, however, the question asks us how the velocity changes. Now, we need to know how kinetic energy and velocity are related. We can pull this relationship from Newtonian mechanics.

\(\displaystyle Kinetic\ Energy=\frac{3}{2}KT=\frac{1}{2}mv^2\)

If we simplify, we can see that \(\displaystyle V=\sqrt{\frac{3KT}{m}}\). Velocity increases with increasing temperature, but in an indirectly proportional manner.

Ideal gases are assumed to have no intermolecular forces and to be composed of particles with no volume. Under high pressure, gas particles are forced closer together and intermolecular forces become a factor. In low temperatures intermolecular forces also increase, since molecules move more slowly, similar to what would occur in a liquid state. Just remember that ideal gas behavior is most closely approximated in conditions that favor gas formation in the first place—heat and low pressure.

Under the ideal gas law, we assume that the interactions between the molecules are very brief and that the forces involved are negligible. The assumption that the molecules obey Coulomb's law when interacting with each other is not necessary; rather, an ideal gas must disregard Coulomb's law.

The ideal gas law assumes only Newtonian mechanics, disregarding any intermolecular or electromagnetic forces.

Measurements of real gases deviate from ideal gas predictions because intermolecular forces and the volume of the particles themselves are not taken into consideration for ideal gases. The volume of the space between particles is considered for ideal gases and does not contribute to deviation from ideal gas behavior.

Attraction between molecules causes real pressure to be slightly less than ideal pressure, while the volume of gas particles causes real volume to be slightly greater than ideal volume.

The question wants you to pick a molecule that will have a negative attraction coefficient, \(\displaystyle a\). The question states that the molecules that repel each other will have a negative \(\displaystyle a\). Recall that similar charges repel each other; therefore, you are looking for an ionized molecule (molecule with a positive or negative charge). In a closed container, these molecules will be force into contact with each other and generate repulsion forces.

Dichloromethane \(\displaystyle (CH_2Cl_2)\) is a neutral molecule. This means that the dichloromethane molecules won’t repel each other. Similarly, sodium chloride is a neutral molecule and will not experience repulsion. Oxygen, with an electron configuration of \(\displaystyle 1s^22s^22p^4\), is also a neutral molecule. If you look at the periodic table, the neutral state of oxygen occurs when there are six valence electrons. The outermost shell of oxygen in this electron configuration \(\displaystyle (n=2)\) has a total of six electrons; therefore, the oxygen has six valence electrons and is neutral.

On the other hand, magnesium, with an electron configuration of \(\displaystyle 1s^22s^22p^6\), is not neutral. Recall that a neutral magnesium atom has two valence electrons and an electron configuration of \(\displaystyle 1s^22s^22p^63s^2\). The magnesium atom in this question, however, has lost two electrons (from the \(\displaystyle 3s\) orbital) and became positively charged with a charge of \(\displaystyle +2\); therefore, the magnesium atoms are ionized, will repel each other, and will have a negative \(\displaystyle a\).

This question can be solved using either Charles's law or the ideal gas law (converted into the combined gas law).

Charles's Law: \(\displaystyle \frac{V_1}{T_1}=\frac{V_2}{T_2}\)

Ideal Gas Law: \(\displaystyle PV=nRT\rightarrow \frac{PV}{T}=nR=\text{constant}\)

The question states that the pressure and moles \(\displaystyle (n)\) are held constant; therefore, the volume and temperature are directly proportional. If the question were asking about an ideal gas, the volume would double when you double the temperature

\(\displaystyle \frac{V_o}{T_o}=\frac{x}{2(T_o)}\rightarrow x=2V_o\)

The volume would double for an ideal gas; however, the question is asking about a real gas. To find the correct relationship between volume and temperature we need to look at the equation for real gas volume. Remember that the volume we are concerned with is the volume of the free space in the container, given by the container volume minus the volume of the gas particles. The equation for real gas volume accounts for the volume of the container and the volume of the gas particles. For a real gas, the volume is given as follows:

\(\displaystyle V_{real}=V_{container}-(n*b)\)

In this equation, \(\displaystyle n\) is the number of moles of gas particles and \(\displaystyle b\) is the bigness coefficient. This equation implies that the volume of free space for a real gas is always less than the volume for an ideal gas; therefore, doubling the temperature will produce a volume that is less than the predicted volume for an ideal gas. Our answer, then, must be less than double the initial volume.

Note that for an ideal gas the bigness coefficient, \(\displaystyle b\), would be zero and the volume of free space \(\displaystyle (V_i_d_e_a_l)\) would be equal to the volume of the container \(\displaystyle (V_c_o_n_t_a_i_n_e_r)\). This occurs because the volume of the gas particles is negligible for an ideal gas.

For a gas to behave ideally, is should have a low mass and/or weak intermolecular forces. Contrastingly, non-ideal gasses should have very large masses and/or have strong intermolecular forces. Therefore, the correct answer is \(\displaystyle HF\) which has very strong intermolecular forces, hydrogen bonds. Nonpolar gasses such as oxygen, and other diatomic gasses have very weak intermolecular forces and thus behave ideally. 

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:

\(\displaystyle P_i=X_iP_{total}\)

In this equation, \(\displaystyle P_i\) is the partial pressure of the gas, \(\displaystyle X_i\) is the mole fraction of the gas, and \(\displaystyle P_t_o_t\) 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.

\(\displaystyle X_i = \frac{n_i}{n_t_o_t}\)

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:

\(\displaystyle X_b=\frac{3mol\ B}{3mol\ A+3mol\ B+6mol\ C}\)

\(\displaystyle X_b = 0.25\)

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

\(\displaystyle P_b = (0.25)(16atm)=4\: atm\)

Similarly, we can find the partial pressure of gas C. The mole fraction of gas C is:
\(\displaystyle X_c=\frac{6mol\ C}{3mol\ A+3mol\ B+6mol\ C}\)

\(\displaystyle X_c=0.5\)

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

\(\displaystyle P_c = (0.5)(16\: atm)=8\:atm\)

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

\(\displaystyle P_c-P_b=8atm-4atm=4atm\)

The partial pressure of gas B is \(\displaystyle 4atm\) less than the partial pressure of gas C.