# Calculation of the work done by an ideal gas

Problem statement:

A sample of ideal gas undergoes a reversible thermodynamic process AB. The volume in the initial state A is VA = 2 m3 and VB = 5 m3 in state B. The relationship between the pressure and the volume in this process is given by the following expression (using SI units): 1. Draw the thermodynamic process in a PV diagram, indicating on the diagram the states A and B as well as the values of p and V for these two states on the corresponding axis.
2. Without calculations, does the ideal gas temperature increase or decrease as it goes from state A to state B?
3. Without calculation, is the work done by the gas positive or negative as it goes from state A to state B? Why?
4. Calculate the work done by the gas as it goes from state A to state B.
5. Use the PV diagram drawn in point a. to graphically calculate the work done by the gas and verify that it equals the work calculated in point d.
6. What are the temperatures of the gas in states A and B if the sample contains n = 10-2 moles?

The gas undergoes a constant-volume process from B to C. The pressure is the same in A and C. Finally the gas undergoes a constant-pressure process from C to A.

1. Draw the ABCA cycle in a PV diagram.
2. What is the work done in process BC? Why? Does the temperature increase of decrease in this process?
3. What is the sign of the work in process CA? Why? Does the temperature increase of decrease in this process?
4. Without calculations, is the work done in process ABCA positive or negative? Why?
5. Use the PV diagram to graphically calculate this work.

Express the results in SI Units.

Givens: R = 8.31 J/mol K

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Solution:

We will use the work done by a gas to solve this problem. The process undergone by the gas as well as states A and B and the isotherms that pass through them are represented in the next figure. Since the gas is ideal, its isotherms are hyperbolas. The gas temperature increases as it goes from A to B, since, as it can be seen in the figure above, state B isotherm is higher than state A isotherm.

With the sign convention used in these pages, the work done by the gas is positive as it expands. We can use the definition of the work done by an ideal gas, substitute the pressure by a function of the volume and use the givens for the volumes to calculate the work. The work is the blue hatched area under the curve which represents the process between state A and B in the figure below: It means that the work is the sum of the rectangular area and the triangle delimited by the process (as shown on the right side of the figure): As expected, the result is the same as the one obtained using the expression of the work done by a gas enclosed in a container.

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We use the equation of state of an ideal gas to calculate the temperature in states A and B: The ABCA process is shown in the figure below: The work done in the process BC is zero since it is an isochoric process and there is no volume change. Temperature in state C is lower than in state B because, as you can see in the upper figure, the (green) isotherm that pass through state C is lower than the one that pass through state B.

The work done in the process CA is negative since the gas compresses. The temperature of the gas decreases; the state A is over an isotherm which is lower than that corresponding to state C.

The work done in process ABCA is positive since the thermodynamic cycle goes clockwise. And it is equal to the area enclosed by the cycle:  The post Calculation of the work done by an ideal gas appeared first on YouPhysics