
The efficiency of a heat engine is the ratio of work output divided by heat input. The process from a to b occurs at constant volume; the heat added, Q_{ab}, therefore depends on the moles of gas, the molar specific heat at constant volume, C_{V}, and the change in temperature. The process from b to c occurs at constant pressure, so the heat added, Q_{bc}, instead depends on the molar specific heat at constant pressure, C_{P}: $$\begin{array}{ccc}{Q}_{ab}=n{C}_{V}({T}_{b}{T}_{a}),& & {Q}_{bc}=n{C}_{P}\left({T}_{c}{T}_{b}\right)\end{array}$$

The equations above and below use the molar specific heat—which you learned about on page 1339. Typical values for air are C_{V} = 20.8 J/(mol K) and C_{P} = 29.1 J/(mol K).

Assume an ideal gas for which PV = nRT. The temperatures at points a, b, c, and d are therefore determined by the pressures and volumes: $$\begin{array}{ccccccc}{T}_{a}=\frac{{P}_{1}{V}_{1}}{nR},& & {T}_{b}=\frac{{P}_{2}{V}_{1}}{nR},& & {T}_{c}=\frac{{P}_{2}{V}_{2}}{nR},& & {T}_{d}=\frac{{P}_{1}{V}_{2}}{nR}.\end{array}$$

After substituting for the temperatures we find the heat inputs Q_{ab} and Q_{bc}: $${Q}_{ab}=\frac{{C}_{v}}{R}{V}_{1}({P}_{2}{P}_{1})\text{}{Q}_{bc}=\frac{{C}_{p}}{R}{P}_{2}({V}_{2}{V}_{1})$$

Consider a large car engine, in which each piston has a 10 cm diameter, stroke of 10 cm, and clearance of 2 cm and operates between 1 and 5 atmospheres (100,000 to 500,000 Pa). Apply the above equations to calculate the input heat as $${Q}_{ab}+{Q}_{bc}=157\text{J}+1,374\text{J}=1,531\text{J}\text{.}$$ The work done in the thermodynamic process is the area inside the PV diagram, or $$({P}_{2}{P}_{1})({V}_{2}{V}_{1})=314\text{J}.$$ The input heat is almost five times the work output of the engine! Furthermore, to keep the pressure constant from b to c requires that the temperature be T_{c} = 15,363ºF! A real engine could not sustain this high a temperature.

This thermodynamic cycle has a terrible efficiency of 21%. Heat not converted to work is rejected to the surroundings during steps c → d and d → a. According to the first law, the heat rejected is the heat added minus the work done, or 1,271 J (79%). Soon we will see how to get better efficiency, but, according to basic thermodynamic principles, no heat engine operating above absolute zero can convert 100% of heat into output work.

 
