According to Boyle's law, pressure and volume are inversely proprotional to each other. This is represented by the equation:

In other words, as volume decreases in a vessel, the pressure will increase, and vice versa. Using the given conditions, we can solve for the final pressure in the vessel:

This question deals with the amount of gas present in a vessel for only one set of conditions. This makes the ideal gas law a suitable equation to use in order to determine the amount of gas in the vessel. The ideal gas law is written as:

Using this equation, we can solve for the molar quantity of gas in the vessel:

Knowing this, we can now solve for the mass of the gas in the vessel by multiplying this molar amount by the molar mass:

Though the ideal gas law gives a nearly close to real approximation of numbers, it oversimplifies its description of gases. No real gas follows the exact definition of the ideal gas law and is very complex because there are intermolecular forces that must be considered. An ideal gas described as a point mass in which the particles are so small that its volume is negligible. However, real gases have real volume. Also, ideal gases are considered elastic, having no attractive and repulsive forces with no energy transfer during collisions. Real gases actually collide and are non-elastic. Note that gases approach ideal behavior as their temperature increases and their pressure decreases.

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.

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

Charles's Law:

Ideal Gas Law: 

The question states that the pressure and moles  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

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:

In this equation,  is the number of moles of gas particles and  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, , would be zero and the volume of free space  would be equal to the volume of the container . This occurs because the volume of the gas particles is negligible for an ideal gas.

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.

Since we're looking at the  orbital, we know . The range of possible values for  is 0 to . Possible values for  range  to . Therefore, among the answer choices,  is the only possible combination of quantum numbers corresponding to an  orbital. 

Since we're looking at the  orbital, we know . The range of possible values for  is 0 to . Possible values for  range  to . Therefore, among the answer choices,

is the only possible combination of quantum numbers corresponding to a  orbital.

If you thought all of these factors are used in the Arrhenius equation, consider that the rate constant is independent of the reactant concentrations. The concentration effect is considered in the rate law by multiplying the concentrations by the constant, but the concentrations themselves are independent of the actual Arrhenius calculation.

The calculation shows that . In this equation, is the activation energy and is the temperature.

Keep in mind that the chemical equation and its coefficients have NOTHING to do with the rate law. The order of each reactant must be determined by experiment.

To find the order of each reactant, compare the initial reaction rates of two trials in which only one of the three reactants' concentrations is altered. For example, trials 1 and 4 keep A and B equal, but C is doubled. When C is doubled, we see that the initial reaction rate is quadrupled. As a result, we determine that the order of reactant C is 2. When the reactant is altered, but the initial reaction rate is kept constant, as seen in trials 1 and 3 with respect to A, the order of that reactant is 0. Finally, when the reactant is multiplied by the same factor that the initial reaction rate is multiplied, as seen in trials 1 and 2 with respect to B, the order of the reactant is 1.

Putting the data together: A is zeroth order, B is first order, and C is second order. Our rate law can thus be written .