# Homework Help: Find the work of this cycle

1. May 5, 2017

### grandpa2390

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

2. Relevant equations
PV = nRT
T= constant

3. The attempt at a solution

P= 103862 Pa,
T = 363.862 K
V = .035 m^3
n = 1.203 moles
at the end.
So the Temperature through the isothermal process should be 362.862

in order to find the work, I need to calculate either the volume or the pressure at the end of the process. I am having trouble figuring out how to do that.

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2. May 5, 2017

### grandpa2390

I thought about using the inverse proportion to work backwards from the end result of the 3rd process to that temperature. and then I would be able to have the other variables filled...
But I don't understand the inverse proportion. it doesn't hold for the end of the 3rd process (when P is 6 bars and V is 10 liters)

3. May 5, 2017

### haruspex

At the end of which stage of the process? You are given the volume at the end of the adiabatic stage.

4. May 5, 2017

### haruspex

It's a matter of fitting the three curves together so that they join up.
For each of the three phases, write an equation for P as a function of V. Put in unknown constants as necessary.
Use the the way that the meet to determine the constants.

5. May 5, 2017

### grandpa2390

The isothermal.

6. May 5, 2017

Why not?

7. May 5, 2017

### grandpa2390

because 6*10^5 is not equal to one divided by .01^2

It would have to be P = 60/v^2

8. May 5, 2017

### Staff: Mentor

Do you understand the difference between a proportional sign and an equal sign?

9. May 5, 2017

### grandpa2390

... you know...
I do, but for some reason, I got it mixed up in my mind to mean "roughly equal to". probably because of the formula. I was expecting it to mean that "1/v^2 is the proportion."
But you are saying that it isn't 1/v^2, it is actually some {proportionality constant} * v^-2

lol. so my approach of working backwards from point a (the initial) to point c (where isothermal meets the third process) is a good start then?

10. May 5, 2017

### Staff: Mentor

Yes.

11. May 6, 2017

### grandpa2390

Here are the values I got. It seems to make sense.

Point C doesn't work out too perfectly because of rounding. perhaps I should use .0165 rather than .017

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12. May 7, 2017

### grandpa2390

I found that the work done by:
a->b is -5912.09 J
b->c is 2733 J
c->a is 2363.64 J

total work is -815.45 J done on the cycle. (or 815.45 J done by the cycle)

13. May 7, 2017

### grandpa2390

Does this make sense?

Q_ab is 0 because it is adiabatic.
Q_bc is -2733 J because it is isothermal. the work done is 2733 J. so -2733 J is the heat absorbed by the cycle
Q_ca is 3548.45 J (heat absorbed).

I got 3548 J because W=-Q and since W = -815 J, then Q = 815 J

so 815 J = -2733 J + Q_ca. so Q_ca absorbs 3548.45 J of heat.

Also, the Q of adiabatic processes = 0. Is this because Q_h = Q_c in an adiabatic process?

I have to calculate the efficiency, and to do that I need to isolate Q_h, and I am wondering if there is Q_h for the adiabatic process that I need to find.

If the adiabatic process absorbs no heat, and the isothermal absorbs no heat. the last process does nothing but absorb heat (T = 1/V from .0165 -> .01, Temperature changes linearly and maxes at the end of the process.)

so e = W/Qh

815.45 J / 3548.45 J = 23% efficient ???

I plugged Q_h = 3548.45 and Q_c = 2733 into the formula $e = 1 - \frac{Q_c}{Q_h}$ and got the same result...
I am not sure if that means much since I used the Wnet = Qnet formula. but it might suggest something about the Q_h and Q_c of each process since I used the formula concerning just the Q_net of each process...

Last edited: May 7, 2017
14. May 7, 2017

### Staff: Mentor

There is no Q for an adiabatic process. By definition, an adiabatic process is one for which Q = 0.
You need to go back and look at process CA in more detail. Part of this process path may involve absorption of heat, and part might involve rejection of heat.

15. May 7, 2017

### grandpa2390

but the temperature rises exponentially. When the process was linear (like on the last problem), the temperature rose to a peak, and then fell somewhere along the process. But in this case, it just rises at T=6/V

$PV = nRT$
$\frac{PV}{nR} = T$
and:
$P = \frac{60}{v^2}$
so:
$T = \frac{60V}{nRV^2}$
$T = \frac{60}{1.203*8.314*V}$
$T = \frac{60}{10V} = \frac{6}{V}$

The temperature only grows with decreasing volume. I thought as long as the temperature was rising, the process would be absorbing heat?
What foolish error am I making this time? : )

16. May 7, 2017

### Staff: Mentor

I analyzed it by determining Q as a function of V, and found that it represents heat addition over the entire path from C to A. So, you're OK.