Calculating Work and Thermal Efficiency of a Heat Engine

[j88k]

Homework Statement

A heat engine using 5.00 g of helium gas is initially at STP. The gas goes through the following closed cycle:
- Isothermal compression until the volume is halved.
- Isobaric expansion until the volume is restored to its initial value.
- Isochoric cooling until the pressure is restored to its initial value.

a)How much work does this engine do per cycle?
b)What is its thermal efficiency?

Homework Equations

thermal efficiency = Wout/QH

(thermal efficiency)Carnot = 1 - TC/TH

The Attempt at a Solution

for part a) I got:

isothermal work = nRTln(V at 2/V at 1) = -1967 J
isobaric work = p(V at 3 - V at 2) = 2836 J
isochoric work = 0

total work done in one cycle = -1967 + 2836 = 869 J

for pat b) i got:

from 1 to 2:
Q = -W = 1967 J

from 2 to 3:
Q = nC_p(deltaT), C_p for a monatomic gas is (5/2)R, this comes from the equipartition theorem
Q = (1.25)(5R/2)(273K) = 7094 J

from 3 to 1:
Q must be negative since there is a work output so therefore Q in this process is irrelevant since we only need the Q that is added to the system

efficiency = (total work done in one cycle)/(heat added in one cycle)
efficiency = (869 J)/(7094 J + 1967 J) = 0.096=9.6% But apparently that's the wrong answer!

I don't know what's wrong. Help!

 

[alphysicist]

Hi j88k,

Homework Statement

A heat engine using 5.00 g of helium gas is initially at STP. The gas goes through the following closed cycle:
- Isothermal compression until the volume is halved.
- Isobaric expansion until the volume is restored to its initial value.
- Isochoric cooling until the pressure is restored to its initial value.

a)How much work does this engine do per cycle?
b)What is its thermal efficiency?

Homework Equations

thermal efficiency = Wout/QH

(thermal efficiency)Carnot = 1 - TC/TH

The Attempt at a Solution

for part a) I got:

isothermal work = nRTln(V at 2/V at 1) = -1967 J
isobaric work = p(V at 3 - V at 2) = 2836 J
isochoric work = 0

total work done in one cycle = -1967 + 2836 = 869 J

for pat b) i got:

from 1 to 2:
Q = -W = 1967 J

Looking at this part: If the gas is being compressed and the temperature is not changing, is heat entering or leaving the system? (In the efficiency formula you only count heat going into the system.) Does this help?

 

[j88k]

so the answer would be (869)/(7094)=0.122=12.2% ?

 

[Redbelly98]

Looks good!