For the gas in Thermo Example 3 undergoing the processes specified there, how much work was done, how much heat was exchanged, and what was the change in internal energy of the gas for each of the four processes?

Given:

Unknown:

Physical Principles and/or Ideas: Behavior and relations for an ideal gas undergoing an isobaric, isothermal, isochoric and adiabatic process, Ideal Gas Law, the First Law of Thermodynamics and relations for heat, work, and internal energy.

Solution: There are four processes in this problem and the heat exchanged; work done and change in internal energy differ for each. However, the internal energy of an ideal gas is what we call a state variable; consequently, we calculate the change in in the same way for all four processes. In contrast, the heat and work are not state variables, which means they are determined by how the gas changes, not just on its initial and final states. So the calculations of the heat exchanged and the work done differ for different processes.

The first process in this problem is an isobaric one. In such a process the gas does work, heat is exchanged, and the internal energy changes. Calculation of the work done is straightforward for isobaric processes, that is:

So for the data we have, we get:

as the work done __by__
the gas. (Notice it is positive since
the gas expands.) But how do we get the
heat exchanged and the change in internal energy?

If we can find one of these two, then we can use the First Law to find the other one. Since the internal energy is a state variable, which depends on only the temperature of the ideal gas, and since we know the number of moles present, we can calculate the change in internal energy from the relation:

Now we apply the First Law to find the heat. We get:

The second process is an isothermal one, which means the internal energy does not change. That is, for an ideal gas, whose internal energy depends on only the temperature, the internal energy remains constant if the temperature does. Consequently, we only need to calculate either the work done or the heat exchanged, and then use the First Law to find the other one. But how do we decide which one to calculate? As is normally the case, we calculate the one for which we have the necessary information. Since we do not have the specific heat, which we would need to calculate the heat exchanged, we will calculate the work. The expression for calculating the work done in an isothermal process is:

Plugging in our values, we get:

Now we use the First Law to find the heat. Since there is no change in the internal energy, the First Law says that the heat exchanged is equal to the work done. So, for this process the heat exchanged is also 3,125 J.

The third process is an adiabatic expansion, which, by definition, means there is no heat exchanged. Consequently, from the First Law we know that the work done is equal to the change in internal energy. Since internal energy is a state variable, and we know the initial and final temperatures, we will calculate the change in internal energy. We get:

It is negative because the internal energy of the gas decreases when the gas does work without any heat input to compensate. Now from the First Law, we have:

So we get:

Notice that the work done by the gas is positive, which it needs to be since the gas expanded.

The final process in the problem is an isochoric one. Since this is a constant volume process, no work is done. The First Law then tells us that the heat exchanged is equal to the change in the internal energy of the gas. Calculating the change in internal energy in the now familiar way, we get:

And this is equal to the heat exchanged.

Notice that the change in the internal energy of the gas was the same for both the isobaric and the isochoric processes since they produced the same final temperatures. This just goes along with what we said about internal energy being a state variable that depends on temperature only. However, the heat exchanged is very different for these two processes. The reason for the difference is that the gas does work in the isobaric process so more heat is needed to reach the same final temperature as in the isochoric process.

Other Thermo Examples: 1 2 3 4 5

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