Question

The working substance of an engine is 1.00 mol of a diatomic ideal gas. The engine operates in a cycle consisting of three steps: (1) an adiabatic expansion from an initial volume of 9.00 L to a pressure of 1.00 atm and a volume of23.4 L, (2) a compression at constant pressure to its original volume of 9.00 L, and (3) heating at constant volume to its original pressure. Find the efficiency of this cycle.

Answer 1

During the adiabatic process there is no exchange of heat , so Q1 = 0

The amount of heat entering or leaving the system during the isobaric compression is

Q2 = Cp nR?T

= (7/2)(1)(P?V)

= (7/2)(1atm)(-23.4 + 9L)

= -50.4 atmL

In adibatic process

PV^? = constant

then P1V1^? = P2V2^?

then P1 = (V2/V1)^? P2

= (23.4/9)^1.4 (1atrn)

= 3.18 atm

The amount of the heat entering or leaving the system during the constant-volume process is

Q3 = Cv nR?T

= (5/2)(1)(?PV)

= (5/2)(6 - 1)(9)

= 112.5 atmL

Therefore the workdone

W = Q1+Q2+Q3

= 0-50.4 + 112.5

= 62.1 atm L

the efficiency

e = 62.1 / 112.5

= 0.552 or 55.2%