# Ideal gas using carnot cycle

1. Jun 1, 2008

### scholio

1. The problem statement, all variables and given/known data
an ideal gas operates in a carnot cycle so that it produces a net positive work of 400 joules per cycle. the maximum temperature during the cycle is 300 deg celsius and the heat lost to a low temperature bath is 600 joules per cycle.

a) what must be the temperature of the low temp bath?

b) what is the change in entropy of the gas during the isothermal expansion

2. Relevant equations

Q = W = nRT ln (Vf/Vi) i don't think is use this equation as moles is not stated, volumes not given either
(entropy) deltaS = deltaQ/T

3. The attempt at a solution

i am not sure what equations to use as i think i need to consider the entire cycle not just the individual isothermic/adiabatic expansions/contractions

will the same W equation from (a) be used in part (b) because in an isothermic expansion Q = W, but again moles and volume is not given?

any help with which equation to use / approach is appreciated...

2. Jun 1, 2008

### konthelion

What is the carnot efficiency?
$$\eta = \frac{W}{Q_{h}}=1-\frac{T_{c}}{T_{h}}$$, W=work, Q=heat

where $$T_{c}$$ and $$T_{h}$$ are in absolute temperature, Kelvins. So you must convert from celsius to Kelvins.

For the second part, you are given that the heat is lost in the cold temperature so use the formula for entropy.
$$s = \frac{Q_{c}}{T_{c}}$$

Last edited: Jun 1, 2008
3. Jun 2, 2008

### scholio

using the equation you specified konthelion

do i sub in 400 for W, 600 for Qh, and 573K = 300 deg cels for Th, then do i solve for Tc?
so

400/600 = 1 -(Tc/573)
573(400/600) = 573 - Tc
-191 = -Tc
191K = Tc

so Tc = -82 deg celsius --> does it sound reasonable?

and for part b, finding the change in entropy of the isothermal expansion using the eq you provided
i got
s = Qc/Tc where Qc = 600joules, and Tc from part a was 191K so s = 6.59 joules/kelvin

reasonable?

cheers

Last edited: Jun 2, 2008
4. Jun 2, 2008

### alphysicist

Hi scholio,

The number for Qh does not look right to me. The problem said that was the heat transferred to the low temperatur bath, so 600 is Qc. How can you find Qh from what is given in the problem? Once you have that, I think the rest of it looks right.

5. Jun 2, 2008

### scholio

i looked through my textbook and found this equation

W/Q_h = (Q_h - Q_c)/Q_h

where W = 400, and Q_c = 600, i solved for Q_h and got 1000

so subsituting 1000 into where i incorrectly entered 600, i solved for T_c and got 343.8K = 70.8 deg celsius

does that sound better now?

also i solved for the change in entropy and got deltaS = Q_c/T_c = 600/343.8 - 1.75 joules/kelvin

6. Jun 2, 2008

### konthelion

Looks correct to me.