Six reversible thermodynamic processes are carried out on one mole of ideal gas in succession to complete a cycle. The processes are two alternate isothermal and adiabatic expansions followed by an isothermal compression and an adiabatic compression to reach to the initial state. The volume changes twofold in the isothermal expansions. Also, the isothermal processes take place at temperatures T1, T2, T3 respectively. The adiabatic exponent of the gas is y.
After the cycle is completed, what is the change in the internal energy of the ideal gas?
No. No need to freak out. The internal energy of the ideal gas depends only on the temperature. After the cycle is completed, the gas returns to its original state – the initial pressure, volume and temperature. That is, after the whole cycle, the temperature does not change and hence, the change in the internal energy ∆E is zero.
This is the consequence of internal energy being a state function.
Draw a rough sketch of the whole cycle on a PV graph, clearly showing the six processes.
Is this a difficult question? No. We do know what the graphs of the two processes are shaped like.
The task left, then, is to draw six curves – three of a definite steepness and three of some other definite steepness – one by one such that they form a closed loop. Consider drawing it yourself, and then take a look at the answer below. While sketching, we needed to keep only one point in mind, what graph would be steeper?
Note : The vertices have been labelled for later use. The direction is ABCDEF initiating at A and terminating at F
Since we know the answer to that question, it is relatively easier for us to draw the curve now. Refer to the point on graphs of adiabatic and isothermal processes
What is the work done by the gas in every cycle? What is the total work done by the gas?
This becomes a fairly easy problem. The reason is that the work done by the gas in the adiabatic process is nothing but -nCv∆T. From the graph, we find that ∆T in each case is T2 - T1, T3 - T2 and T1 - T3 respectively. Cv is R/y-1. So, the adiabatic processes are dealt with.
Work in isothermal processesEdit
The task reduces to finding the work done in the isothermal processes. Make a mental note that the isothermal processes perform a work –nCvln(V2/V1). Do we know each of the value substitutions? Yes. We know n, Cv, the temperatures at which the isothermal processes take place, and the volume ratio = 2 for the expansions. The only unknown is the volume ratio to be substituted inside ln in the isothermal compression. What is this ratio? And how are we to find it?
Finding the required volume ratioEdit
Recall that for an unknown, we need to create an equation. And most of the times, an equation is created because some variables are restricted to some values by a certain law.
Just remember that the initial and final volumes and temperatures (and all intermediate states) in an adiabatic process are constrained to satisfy T1V1y-1 = T2V2y-1. This is the constraint we were talking about.
Let us mark the vertices of this curve as A, B, C, D, E, F respectively, A being the initial point of the T1 isotherm. Then, what we are to find is the ratio VE/ VF.
Here are the three equations: -
- VB/VC = (T2/T1)^[1/(y-1)]
- VD/VE = (T3/T2)^[1/(y-1)]
- VF/VA = (T1/T3)^[1/(y-1)]
Multiplying all three equations, we get the required ratio to be 4. (The RHS after multiplication will reduce to 1, always make a mental note of such cyclic expressions. On the LHS, you will encounter two known ratios an substitute their value as 2.)
- WAB = -nRT1ln2
- WBC = -n[R/(y-1)](T2 - T1)
- WCD = -nRT2ln2
- WDE = -n[R/(y-1)](T3 - T2)
- WEF = -nRT3ln0.25 (Because this is a compression)
- WFA = -n[R/(y-1)](T1 - T3)
Since Cv was not known, we had to express it in the terms of the known adiabatic exponent y. The value of n is one mole, but it has been reported to keep harmony with dimensional formulae.
- Found the work done in adiabatic prcess as –nCv∆T
- Found the work done in isothermal process as -nRT ln(V2/V1)
- Found the ratio V2/V1 for the isothermal compression using the condition of reversible adiabatic processes to generate three equations, and multiplying them.
Find the heat exchange of the ideal gas in each step.
This is a mechanical step. All you have to do is use the first law of thermodynamics.
- QAB = nRT1ln2
- QBC = 0
- QCD = nRT2ln2
- QDE = 0
- QEF = nRT3ln0.25 = -2nrtT3ln2
- QFA = 0
What is the efficiency of the process?
Now, the efficiency of the process has to be thought of as the ratio of total output and total input. In thermodynamic processes, only the net work done (∫PextdV over all the processes, not the IUPAC w) qualifies as output. We will NOT consider heat rejected by the system as output.
The input, on the other hand, is not the net heat exchange. It is rather the heat given to the system. The heat rejected by the system does not occur anywhere in the numerator or the denominator.
Then, the output is nR(T1 + T2 - 2T3)ln2. The input is nothing but QAB + QCD = nR(T1 + T2)ln2.
We did not count QEF because it was the heat rejected by the system.
The final answer is the ratio