# Reversible cycle with an adiabatic process (ideal gas)

Problem Statement:

A sample of monatomic ideal gas is initially at temperature T1 = 400 K, pressure p1 = 5 atm and occupies a volume V1 = 0.6 m3. The gas expands isothermally to a volume V2 = 3 m3. Its pressure then drops at constant volume up to a state 3. Finally the gas returns to its original state after completing the cycle with an adiabatic compression. All the processes are reversible.

1. Draw the thermodynamic cycle in a PV diagram.
2. Calculate the pressure in state 3 using the adiabatic process equation (between states 3 and 1). What is the temperature in this state?
3. Calculate the work done, the internal energy change and the heat exchanged by the gas for each process and over the entire cycle.

Express the results in SI units.

Givens: R = 8.31 J/mol K; CV = (3/2)R 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. We recommend you to give a look at the reversible processes of an ideal gas page before trying to solve it.

The thermodynamic cycle undergone by the ideal gas is represented in the next figure: The equation for an adiabatic process expressed as a function of pressure and volume is: Where γ is the heat capacity ratio (or adiabatic index), that in the case of a monatomic ideal gas is given by: Pressure and volume in state 1 are knowns; on the other hand, the volumes in states 3 and 2 are the same (see the thermodynamic cycle in the PV diagram), so we can use the adiabatic process equation to calculate the pressure in state 3: Where we have expressed the pressure in pascals using the following conversion: We can use the equation of state of an ideal gas applied to state 3 to calculate the temperature. But first we need to determine the number of moles in the sample of ideal gas. And we can determine it by applying the equation of state of an ideal gas to state 1: We can now use the equation of state of an ideal gas applied to state 3: As expected, the temperature in state 3 is lower than in state 2, since from state 2 to state 3
the pressure decreases at constant volume.

Keep in mind that the numerical results will be slightly different depending on the amount of decimals used to do the calculations. Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!

We will now calculate the work done, the heat exchanged and the internal energy change for each process.

Isothermal process 1-2: Since the gas expands, the work done by the gas and the heat exchanged are both positive. Isochore process 2-3: The gas gives off heat and its internal energy decreases since its temperature falls. Adiabatic process 3-1: During this adiabatic compression, the internal energy increases since the temperature rises. Furthermore, since it is a compression, the work done by the gas is negative. Finally, the heat and the internal energy change over the entire cycle are respectively: Since the cycle goes clockwise, the work done by the gas is positive. There is no internal energy change during the cycle because it is a state function. Therefore, the heat exchanged equals the work done by the gas.

The work done by the gas over the entire cycle is the blue area enclosed by the cycle: The post Reversible cycle with an adiabatic process (ideal gas) appeared first on YouPhysics