A cyclic process is a series of transformations that take the gas back to its original state. These form a closed loop on a \(pV\) diagram.
Key Concepts
A simple cyclic process involves:
- Isobaric expansion - work is done on the surroundings
- Isovolumetric cooling
- Isobaric compression - work is done on the gas
- Isovolumetric heating
Let's consider the cycle for a heat engine.
Since \(W=p\Delta V\), the area between a line on the \(pV\) diagram and the \(V\) axis gives work done. The net work done during a cyclic process is given by the enclosed area.
We can show the net work done using a Sankey diagram.
The dimensionless quantity, thermal efficiency, of any mechanical process is defined as the ratio of the useful work done to the total energy input:
\(\eta={\textrm{useful work done}\over \textrm{energy input}}\)
For a heat engine:
- Energy input is \(Q_\textrm{H}\)
- Work done is \(W_\textrm{out}=Q_\textrm{H}-Q_\textrm{C}\)
\(\eta={Q_\textrm{H}-Q_\textrm{C}\over Q_\textrm{H}}=1-{Q_\textrm{C}\over Q_\textrm{H}}\)
The most efficient cyclic process is known as the Carnot cycle. This involves:
Isothermal expansion - work is done on the surroundings- Adiabatic expansion - work is done on the surroundings
- Isothermal compression - work is done on the gas
- Adiabatic compression - work is done on the gas
Note that you may be required to calculate the work done during any part of the process using the 'counting squares' method. This is a little arduous!
- Calculate the size represented by one square using the bottom left square on the axes
- Count the squares enclosed in the area required and multiply
To calculate the thermal efficiency we must consider entropy.
The total change in entropy, \(\Delta S=\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4\)
Since stages 2 and 4 are adiabatic, \(Q=0\) and so \(\Delta S _2 = \Delta S_4=0\)
Stages 1 and 3 take place at constant temperature, so \(\Delta S_1 = {Q_\textrm{H} \over T_\textrm{H}}\) and \(\Delta S_3 = {Q_\textrm{C} \over T_\textrm{C}}\)
Since the Carnot cycle is a reversible process, \(\Delta S = {Q_\textrm{H} \over T_\textrm{H}} - {Q_\textrm{C} \over T_\textrm{C}} = 0\)
\( {Q_\textrm{H} \over T_\textrm{H}} = {Q_\textrm{C} \over T_\textrm{C}} \Rightarrow {Q_\textrm{C} \over Q_\textrm{H}} = {T_\textrm{C} \over T_\textrm{H}}\)
Substituting into the general equation for thermal efficiency of a heat engine:
\(\eta_\textrm{Carnot}=1-{T_\textrm{cold}\over T_\textrm{hot}}\)
The equation for thermal efficiency for a Carnot cycle makes clear that we can approach but never attain 100% efficiency. To do so we would need to have a high temperature thermal energy store approaching infinity or a low temperature thermal energy sink approaching absolute zero:
- If \(T_\textrm{hot} \rightarrow \infty\), \(\eta \rightarrow1\)
- If \(T_\textrm{cold}\rightarrow 0 \textrm{K}\), \(\eta \rightarrow1\)
At the operating temperatures provided, the Carnot cycle is the most efficient process possible.
Use flashcards to practise your recall.
Use quizzes to practise application of theory.
START QUIZ!
This \(pV\)diagram shows a cyclic process. The amount of gas present is such that \(nR = 10\) kPa cm3K-1 \(=0.01\) Pa m3K-1.
Calculate the temperature change during the transition from A to B.
Using the ideal gas law, calculate temperature separately at each point \(T = {PV\over nR}\):
- \(T_A = 2000\text{ K}\)
- \(T_B = 4000\text{ K}\)
This \(pV\) diagram shows a cyclic process. The amount of gas present is such that \(nR = 10\) kPa cm3K-1 \(=0.01\) Pa m3K-1.
Calculate the internal energy change during the transition from A to B.
The difference in temperature (calculated using the ideal gas law for the two points) is 2000 K.
\(ΔU = {3\over 2}nRΔT = {3\over 2} \times 0.01 \times 2000\)
This \(pV\) diagram shows a cyclic process. The amount of gas present is such that \(nR = 10\) kPa cm3K-1 \(=0.01\) Pa m3K-1.
Calculate the work done by the gas during the transition from A to B.
The work done is equal to the area under the graph: \(W = 200 \times 10^3 \times 100 \times 10^{-6}\)
This \(pV\) diagram shows a cyclic process. The amount of gas present is such that \(nR = 10\) kPa cm3K-1 \(=0.01\) Pa m3K-1.
Calculate the work done by the gas during the transition from B to C.
There is no change in volume, which means there has been no displacement. Nor is there any area enclosed under a vertical straight line. Therefore, no work has been done.
This \(pV\) diagram shows a cyclic process. The amount of gas present is such that \(nR = 10\) kPa cm3K-1 \(=0.01\) Pa m3K-1.
Calculate the heat removed from the gas during the transition from B to C.
Using the ideal gas law, calculate temperature separately at each point \(T = {PV\over nR}\):
- \(T_B = 4000\text{ K}\)
- \(T_C = 2000\text{ K}\)
\(Q = ΔU = {3\over2}nRΔT = {3\over2} \times 0.01 \times (-2000)\)
We are asked for the heat removed from the gas: \(Q=+30\text{ J}\)
This \(pV\) diagram shows a cyclic process. The amount of gas present is such that \(nR = 10\) kPa cm3K-1 \(=0.01\) Pa m3K-1.
Calculate the work done on the gas during the transition from C to D.
The work done is equal to the area under the graph: \(W = 100 \times 10^3 \times (-100) \times 10^{-6}\)
Since we are asked for work done on the gas: \(W=+10\text { J}\)
This \(pV\) diagram shows a cyclic process. The amount of gas present is such that \(nR = 10\) kPa cm3K-1 \(=0.01\) Pa m3K-1.
The net work done by the gas in the cycle is:
Net work done by the gas is the area enclosed (in a clockwise direction):
\(W = 100 \times 10^3 \times 100 \times 10^{-6}\)
This \(pV\) diagram shows a Carnot cycle. The amount of gas present is such that \(nR = 10\) kPa cm3K-1 \(=0.01\) Pa m3K-1.
The temperature is maximum at:
The transition from A to B is a high temperature isotherm. This isotherm is always at higher temperature than the third stage because more work needs to be done by the gas than on it.
NOTE: B and C (and A and D) are at different temperatures as these stages are adiabatic.
This \(pV\) diagram shows a Carnot cycle. The amount of gas present is such that \(nR = 10\) kPa cm3K-1 \(=0.01\) Pa m3K-1.
The net work done by the gas is approximately:
The area enclosed by the cycle is about one square. Each square represents \(100\times 10^3\times100\times 10^{-6}\)J.
HINT: It's easiest to calculate the work done equivalent for one square by looking at the data for the square at the origin.
This \(pV\) diagram shows a Carnot cycle. The amount of gas present is such that \(nR = 10\) kPa cm3K-1 \(=0.01\) Pa m3K-1.
Most work is done by the gas from...
The area between the graph and the horiztonal axis is largest under stage A to B.
How much of Cyclic processes have you understood?
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