# Thermodynamics (Heat cycles)

1. Feb 3, 2010

### kd001

1. The problem statement, all variables and given/known data

One mole of an ideal monatamic gas at an initial volume of 25L follows the cycle shown below. All the processes are quasistatic. Find the efficiency of the cycle.

2. Relevant equations

PV = nRT
dQ (at constant volume) = Cv*dT
dQ (at constant pressure) = Cp*dT
W(isothermal) =nRTln(Vi/Vf)

3. The attempt at a solution

Firstly I calculated the temperature at each stage using the first equation. I found the first temperature to be 301K and the next two to be 601K.

For the constant volume and the constant pressure stages I calculated the heat flow using the second and third equations. I found Q12 to be 3700J, and Q31 to be -6200J.

For the isothermal portion heat flow is equal to work done by the gas which I found to be 3500J (so 3500J is flowing into the gas and the gas is doing 3500J of work).

For the constant volume portion the work done is 0 and for the constant pressure volume the work done on the gas is pressure times change in volume (-2500J).

To calculate efficiency I divided the net work done by the gas by the net heat flow into the gas but the answer is 99.9%. Can anyone spot where I have gone wrong?

Thanks a lot.

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2. Feb 3, 2010

### Andrew Mason

I am not sure how you got 99.9%. The net work is the area inside the path. That is 1000 J. (3500-2500 but which is also Qh-Qc = 7200-6200 = 1000). The efficiency is:

$$\eta = W/Q_h = 1000/7200 = 13.9$$%

If you are using the net heat flow, Qh-Qc, in the denominator you will always get 100% since W = Qh-Qc.

AM

3. Feb 3, 2010

### kd001

Ok. I now know where I went wrong. I got the same value for W but I divided it by the net heat flowing into the system (ie 3700 + 3500 - 6200) instead of Qh.