In this question, let's start with what we know and what we want to know. We're given the volume of the gas in a container, as well as the pressure and the temperature. Then, we're asked to find the number of moles of gas that are in this container.

To answer this question, we'll need to make use of the ideal gas equation.

Plugging in the values that we know, we can calculate our answer.

Using

Solving for

Plugging in values

Using ideal gas law:

Converting Celsius to Kelvin and plugging in values:

We will use the ideal gas law to solve this problem:

We are asked to find the change in volume, so let's rearrange for that:

Then applying this to the initial and final scenarios:

Then taking the ratio of scenario 2 to scenario 1:

Where:

Substituting these in, we get:

We will begin with the ideal gas law for this problem:

Then rearranging for total moles:

Then multiplying by 80% to get the number of moles of nitrogen:

To determine the answer, we must do several steps with the ideal gas law:

For this problem, the pressure and temperature are the only things to change. Therefore, we can rearrange the equation to account for this.

However, we don't know what the value of P1 is just yet. To determine it, we must use the values we do know to solve for the initial pressure.

Now, we have P1. Using this, along with T1 and T2, we can solve for P2.

In this question, we're given a situation in which a gas with certain defined parameters is undergoing an isobaric expansion. We're asked to determine how the temperature changes.

An isobaric expansion refers to a process in which a container expands while maintaining a constant pressure. In this problem, we're going to need to use the ideal gas law.

Based on the process described in the question stem, we know that the number of moles of gas will remain the same. Furthermore, we know the pressure remains constant. And, of course, the ideal gas constant stays the same as well. Thus, we can lump all the constants together.

Therefore, the only variables that will be changing during this process are the temperature and the volume. Consequently, we can set up a before and after expression.

Since we're solving for the final temperature, , let's go ahead and isolate that term.

Since we know the volume doubles, we can conclude that . We can plug this into the equation to obtain:

Canceling common terms, we obtain:

Thus, we see that the final temperature is twice as much as the initial temperature. We could have concluded this fairly easily without the math, since we know that temperature and volume are directly proportional to one another. If one doubles, the other will also double (assuming all other variables remain constant).

Using

So

Solving for 

Converting to Kelvin and plugging in values

For this question, we're given various parameters with regards to a gas, including its pressure, temperature, and density. We're asked to solve for the gas's molar mass.

To solve this problem, we'll need to make use of the ideal gas law.

Using this expression, we'll need to manipulate it in order to account for the density of the gas. To do that, we'll have to define moles in an alternative way.

Combining these two expressions....

And rearranging further....

Leads us to a term for density.

Now, we can isolate the term for molar mass.

Then we just need to plug in the values from the question stem to find the molar mass.

This question is asking us to determine a true statement regarding entropy. Let's look at each answer choice to see what is true about entropy and what isn't.

• The entropy of a system, whether it is isolated or non-isolated, can only increase.

This above statement is not true. While it's true that the entropy in an isolated system can only increase, the entropy in a non-isolated system can either increase or decrease. However, for the entropy to decrease in a non-isolated system, the entropy of the surroundings needs to increase by a greater amount.

• The entropy of a system can decrease, but only if the system is isolated and the process is irreversible.

This is another false statement. Once again, the entropy of an isolated system cannot decrease; it can only increase. Furthermore, only irreversible processes will result in an increase of entropy in such systems.

• The entropy of a system can decrease only in the case of a reversible adiabatic process.

Again, this is another false statement. The change in entropy of a system that is associated with a truly reversible process can be shown mathematically by the following equation:

This equation shows that in a reversible process in which an infinitesimal amount of heat is added to (or taken away from) a system at a given temperature, the change in entropy of that system can be calculated. There are no truly reversible processes that occur in nature, as such a process would take an infinite amount of time.

In an adiabatic process, there is no heat transfer. Thus, the  term in the above equation is equal to . Consequently, there is no change in the entropy of the system.

• The entropy of a non-isolated system can decrease only if the entropy of its surroundings increases by a greater amount.

This is a true statement. In any non-isolated system, such as a refrigerator, the entropy can certainly decrease. However, since all irreversible processes must result in an increase in entropy in the universe as a whole (second law of thermodynamics), the entropy of the surroundings must decrease. We can express this mathematically as:

As can be seen by the above equation, if the  term is negative, then the  term must not only be positive, but it must also be of greater magnitude.