# Thermal efficiency of the engine in a PV diagram

#### YC_128

A heat engine takes 2.0 moles of an ideal gas through the reversible cycle abca, on the pV diagram shown in the figure. The path bc is an isothermal process. The temperature at c is 820 K, and the volumes at a and c are 0.010 m3 and 0.16 m3, respectively. The molar heat capacity at constant volume, of the gas, is 37 J/mol·K. The thermal efficiency of the engine = (net workdone on the system) /(total heat input to the system) , is close to? (image: https://postimg.cc/QF6mj9db )

1. Homework Statement

n=2 mol
Tb=Tc=820K (isothermal)
Va=0.010m3
Vc=0.16m3
Cv=37 J/mol·K

2. Homework Equations
Work done=nRT*ln(V2/V1)
Work done=P(V2-V1)
PV=nRT
e=(Work done)/(Qheat)

3. The Attempt at a Solution
I
first find the work done by Wbc-Wca

Wbc=nRT*ln(Vc/Va)=37785.94814J

Wca=P(0.16-0.01) (Where P is constant through c to a Pc=Pa PcVc=nRT Pa=Pc=85177.5Pa)

Wca=12776.624J

Wbca=Wbc-Wca=25009.32314J

PaVa=nRTa Ta=(nR)/(PaVa)=(2*8.31)/(85177.5*0.010)=51.25

Qheat=Qab=nCv(Tb-Ta)=2*37*(820-51.25) =56887.5J

e=Wbca/Qab=25009.32314/56887.5=0.439627741 which is wrong

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#### Chestermiller

Mentor
What about the heat added in bc?

#### kuruman

Homework Helper
Gold Member
It might be easier and more educational to organize your thoughts with problems of this sort by filling the blanks in a Table as shown below without using the first law of thermodynamics (figuring out what's left over.) Then you can check your work by verifying that the entries in any row satisfy the first law. After the Table is correctly completed, one can answer all questions related to the problem. For example the efficiency of the cycle is the sum of all the entries in column $W$ divided by the positive entries in column $Q$. The starting point is figuring out the temperatures at the vertices and using them to fill in column $\Delta U$.
$$\begin{array}{|c|c|c|c|} \hline & \Delta U~~ & Q~~~~~ & W~~~~ \\ \hline a \rightarrow b & & & \\ \hline b \rightarrow c & & & \\ \hline c \rightarrow a & & & \\ \hline \end{array}$$

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