**Problem Statement:**

Two moles of an ideal monatomic gas undergo a reversible cyclic process ABCA as shown in the figure below. The temperature in state A is T_{A} = 400K and it is T_{B} = 300K in state B. The work done by the gas during the process CA is W_{CA} = 6000 J.

- Is the cycle total work output positive, negative or zero? Same question for the total heat exchanged. Why?
- Calculate the change in the internal energy for each step and over the entire cycle.
- Calculate the heat exchanged for each step of the process and over the entire cycle.
- Calculate the work done by the gas for each step and over the entire cycle.

Express the results using SI units.

__Givens__: R = 8.31 J/mol K; C_{V} = (3/2)R

Ad blocker detected

Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!

**Solution:**

This problem is an application of the First Law of Thermodynamics.

Since the cycle goes clockwise the work done by the gas on its surroundings is positive.

The **change in the internal energy over the entire cycle is zero because the internal energy is a state function**. So, applying the First Law, we get that the heat exchanged during the cycle equals the net work:

And since this work is positive, the net heat exchanged by the gas is positive as well, thus absorbed.

The change in the internal energy of the ideal gas (independently of the process undergone) is given by:

Where C_{V} is the ideal gas molar heat capacity at constant volume.

You can see in the problem statement PV diagram that the gas temperatures are the same for states A and C, as they are located on the isotherm at temperature T_{A}. Therefore the internal energy change for process CA is zero:

Furthermore, since the total cycle internal energy change is zero, we will have:

After substituting the data we obtain:

As we saw before, the total internal energy change over the entire cycle is zero.

Ad blocker detected

Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!

The work done during process CA is W_{CA} = 6000 J and the internal energy change is zero. Therefore, by applying the First Law, we get that the heat exchanged during the process is equal to the work done by the gas:

The work done during process BC is zero because there is no volume change. Therefore, by applying the First Law, we get:

Process AB is isobaric, therefore the heat exchanged during this process is given by:

Where C_{p} is the ideal gas molar heat capacity at constant pressure that we can deduce from C_{V} using Mayer’s relation.

We use the First Law of Thermodynamics to calculate the work done by the gas in process AB:

That is negative as expected because the gas compresses in process AB.

Finally, the heat, the work and the internal energy change over the entire cycle are given by:

The post Application of the First Law of Thermodynamics appeared first on YouPhysics