Ideal gases have particles that are considered to be point masses. This means that all of their (extremely small) mass is contained in a single, infinitesimal point of zero volume. These particles experience elastic collisions, which means they lose no kinetic energy when they collide with other particles. If the particles experienced strong intermolecular forces, then they couldn't have elastic collisions, so they wouldn't be ideal gases anymore. Ideal gases don't experience any intermolecular forces.

When the temperature is higher, the kinetic energy of particles is higher. Because their energy is higher, they bounce more energetically, letting them ignore most intermolecular forces (which is something ideal gases don't experience). 

An important assumption for ideal gases is that the volume of the particles is negligible compared to the volume of their container. For real gases, as pressure increases, the particles get closer together, and their volume gets less negligible when compared to the volume of the container. This is why real gases behave more ideally at lower pressures.

Use the ideal gas law equation.

$\displaystyle PV=nRT$

Manipulate the equation, substitute the givens, and solve for temperature. Write the gas constants necessary to solve the problem.

$\displaystyle T=\frac{PV}{nR}$

$\displaystyle 1{ atm}= 101.3{ kPa}= 101325 \frac{{N}}{{m}^2}$

$\displaystyle R= 8.314{ }\frac{{L}\cdot {kPa}}{{mol} \cdot {K}}$

$\displaystyle T=\frac{PV}{nR}= \frac{(5{ atm})(\frac{101.3{ kPa}}{1{ atm}})(10{ L})}{(1{ mol})(8.314{ }\frac{{L}\cdot {kPa}}{{mol} \cdot {K}})}$

$\displaystyle T=609.213 { K}$

An ideal gas has no molecular interactions besides perfectly elastic collisions. The energy is the combined kinetic energies of ideal gas molecules since there is no potential energy. For $\displaystyle N$ point molecules of the monatomic gas, the equation can be written as:

$\displaystyle U=\frac{3}{2}NkT$

Note that $\displaystyle K_{{avg}}=\frac{3}{2}kT$ is the equation for average kinetic energy also in Joules.

We will use the relationship:

$\displaystyle \frac{V_1}{T_1}=\frac{V_2}{T_2}$

Rearrange for the final volume: 

$\displaystyle \frac{V_1 T_2}{T_1}={V_2}$

Convert our temperatures to kelvin. 

$\displaystyle 19^{\circ}C+273=292^{\circ}K$

$\displaystyle 38^{\circ}C+273=311^{\circ}K$

Plug in our values and solve.

$\displaystyle \frac{V_1 T_2}{T_1}={V_2}$

$\displaystyle \frac{10L* 292^{\circ}K}{311^{\circ}K}={V_2}$

$\displaystyle 10.65 L=V_2$

We will use the relationship:

$\displaystyle \frac{V_1}{T_1}=\frac{V_2}{T_2}$

Rearrange to solve for the final volume:

$\displaystyle \frac{V_1 T_2}{T_1}={V_2}$

Convert our temperatures to kelvin: 

$\displaystyle 10^{\circ}C+273=283^{\circ}K$

$\displaystyle 30^{\circ}C+273=303^{\circ}K$

Plug in our values and solve.

$\displaystyle \frac{15L* 303^{\circ}K}{283^{\circ}K}={V_2}$

 $\displaystyle 16.1 L=V_2$

We will use the relationship:

$\displaystyle P_1V_1=P_2V_2$

Rearrange to solve for the final volume: 

$\displaystyle \frac{P_1V_1}{P_2}=V_2$

Plug in our values and solve.

$\displaystyle \frac{1.3atm*13L}{.85atm}=V_2$ 

$\displaystyle 19.9 L=V_2$

We will use the relationship:

$\displaystyle P_1V_1=P_2V_2$

Rearrange to solve for the final volume:

$\displaystyle \frac{P_1V_1}{P_2}=V_2$

Plug in our values and solve. 

$\displaystyle \frac{.5atm*6L}{1.5atm}=V_2$ 

$\displaystyle 2L=V_2$

First, we will convert celsius to Kelvin.

$\displaystyle 10^{\circ}C+273=283^{\circ}K$

$\displaystyle 50^{\circ}C+273=323^{\circ}K$ 

Then, we will use the relationship:

$\displaystyle \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}$

Solve for volume final volume:

$\displaystyle \frac{P_1 V_1 T_2}{T_1 P_2}=V_2$

Plug in our values and solve. 

$\displaystyle \frac{1atm* 16L*323^{\circ}K}{283^{\circ}K *5atm}=V_2$

 $\displaystyle 3.65 L=V_2$

When talking about gases, it's convenient to talk about gases that behave ideally. The ideal gas law, $\displaystyle PV=nRT$, generally only holds true for gases that behave ideally. Once gases begin to deviate from ideal behavior, this equation begins to lose its accuracy in its power to predict.

There are a number of assumptions that are made when referring to gases that behave ideally, and these assumptions constitute what is known as the kinetic molecular theory of gases. In this model, there are a total of 5 assumptions:

1. Individual gas particles are very tiny points of mass that essentially have no volume (i.e. the volume of a gas is equal to the volume of free space in the container).

2. These gas particles are neither attracted to, nor repelled by, other gas particles (i.e. there are no intermolecular forces acting between gas particles).

3. Each of these gas particles are continuously moving in random directions.

4. All of these gas particles collide with each other, such that no energy is gained or lost by each collision (i.e. each collision is completely elastic).

5. At a given temperature, the average kinetic energy for a given sample of gas is the same as any gas (i.e. the identity of the gas does not matter) and the average kinetic energy of the sample of gas is directly related to the absolute temperature.

Now that we know what the assumptions are, let's go ahead and see how pressure and temperature affect ideal behavior, since these two parameters show up in each of the answer choices.

Temperature

When the temperature of a gas is lowered, the average kinetic energy of the sample of gas also decreases. Consequently, each gas particle making up the solution is expected to slow down. When this happens, the gas particles become more available for interacting with other gas particles, which increases the likelihood for intermolecular forces to begin playing a role. This violates one of the assumptions of kinetic molecular theory. When individual gas particles begin to attract one another, the pressure that the gas exerts on the wall of its container becomes less than what the ideal gas law equation would predict.

Thus, we would expect a high temperature to cause the least deviation from ideal behavior. In other words, a high temperature would increase the average kinetic energy of the sample of gas, and when all of the gas particles are shooting around at greater speeds, they have less of an opportunity to interact with other gas particles.

Pressure

When the pressure of a gas is increased, we also see deviations from ideal behavior. Remember that one of the assumptions of kinetic molecular theory is that the individual gas particles are very tiny points of mass that essentially have no volume, which means that the volume of the gas as predicted by the ideal gas law is essentially equal to the volume of the container. But, we of course know that real gases do not have zero volume and instead, each individual gas particle takes up a small amount of volume. As the external pressure becomes greater and greater, the volume of the container decreases as well. This, in turn, causes each gas particle to comprise a greater fraction of the container. Consequently, at high enough pressures, a real gas will take up a significant fraction of the container's volume, meaning that the volume of the real gas will be greater than what is predicted by the ideal gas law.

When external pressures are kept low, the volume of the container is sufficiently large enough that the volume of gas takes up a negligible fraction of the container. Thus, low external pressures will be expected to cause the least deviation from the ideal gas law.

In summary, it's important to remember the following:

• At high pressures, the actual volume of gas particles makes the real volume higher than the predicted volume.
• At low temperatures, intermolecular forces make the actual pressure lower than the predicted pressure.