# Homework Help: 2 Thermodynamic questions

1. Mar 25, 2008

### ~christina~

[SOLVED] 2 Thermodynamic questions

1. The problem statement, all variables and given/known data
1.
a) Determine the work done on a fluid that expands from i to f as indicated in picture below.
b) How much work is performed on the fluid if it is compressed from f to i along the same path?

http://img340.imageshack.us/img340/6344/82457490fq4.jpg [Broken]

2. An ideal gas initially at Pi, Vi and Ti is taken through a cycle as shown.
a) find the net work done on the gas per cycle
b) What is the net energy added by heat to the system per cyle
c) Obtain a numerical value for the net work done per cycle for 1.00 mol of gas initially at 0C

2. Relevant equations
$$\Delta E_{int}= Q+ W$$
3. The attempt at a solution

1.

a)Determine the work done on a fluid
I thought that work was the area under the curve...but acording to my book for constant pressure.. I think it is
$$P_i(V_f-V_i)$$

Now I'm thinking that If I find the are under the curve then it would be=>
(4-1m^3)(2x10^6Pa)+ (2-1m^3)(6x10^6-2x10^6)+ 1/2(3-2m^3)(6x10^6-2x10^6)=
(6x10^6)+(4x10^6)+ (2x10^6)= 1.2x10^7 J

b) How much work is performed on the fluid if it is compressed from f to i along the same path?
here I'm not sure what they want, when they say that the fluid is compressed along the same path. I assume that would mean that the pressure changes? But how would that look on the graph? Am I supposed to draw a new graph?

_________________________________________________________________________
2.

a) net work done on the gas per cycle
I would say work done is = area of the cyclic cycle thus:

(3Pi-Pi)(3Vi-Vi)= 4J

b) What is the net energy added by heat to the system per cyle

if it is cyclic I think that the net energy added by heat would be =
Q= -W because the $$E_{int}= 0$$ thus would it would be -4J ?

c) Obtain a numerical value for the net work done per cycle for 1.00 mol of gas initially at 0C

Not sure how I would do this if they say that there is 1.00mol of gas at 0C.
I know that $$Q=nc\Delta T$$ but if you don't have the final temp then how do you do the question, and also if you don't have c?

I need help on kowing whether I'm correct or not and need help on parts c for #2 and also on part b) of #1.

Thank you.

Last edited by a moderator: May 3, 2017
2. Mar 25, 2008

### ~christina~

um..does anyone know how to do thermodynamic Q's?
not so sure b/c I haven't seen alot of people ask them...

Well If anyone is capable and willing to help me out I'd greatly appreciate it

3. Mar 25, 2008

### Andrew Mason

Your first thought is correct. Pi(Vf-Vi) IS the area under a constant pressure PV graph. But in this case, you do not have constant pressure.

Read the question carefully. You are asked for the work done ON the fluid. Is positive work being done on the fluid here?

Again, this is a matter of a +- sign. How does your answer differ from that in a)?

_________________________________________________________________________
Be careful about the sign. The question asks for the work done ON the gas. Is net positive work done on the gas during a cycle?

That would make the system 100% efficient, which is not possible.

When does heat flow into the system? (ie. does heat flow into or out of the system in each of AB, BC, CD, DA?). How much heat flows into/out of in each of these segments (use the first law: dQ = dU + W = dU + PdV ). Add them up (using correct signs) and that gives you the net heat flow.

What is the final temperature after completing one full cycle? You can determine the temperature at each point using the ideal gas law.

AM

4. Mar 25, 2008

### ~christina~

no it is negative

It is possitive since the gas is compressed thus volume decreases instead of increases.

_________________________________________________________________________
I'm not sure but I say no but if
WAB and WBC is = 0 then

WBC = 3Pi(3Vi-Vi) and the same is for the WAD just negative then wouldn't it = 0??

$$Q=Mc\Delta T$$
AM[/QUOTE]

Thank You

5. Mar 25, 2008

### Andrew Mason

You had the right idea - the work done by the gas is the area of the rectangle 2Pi x 2Vi. So the work done ON the gas is - 4PiVi. You can't convert this to Joules unless you know how Pi and Vi.
You can determine the temperature at each point using the ideal gas law. The Q from A to B is simple since there is no work done: dQ = dU. What is dU for AB? ($\Delta U = nC_v\Delta T; \Delta T = \Delta (PV)/nR = V_i\Delta P/nR$). You need to know the kind of gas (diatomic, monatomic) in order to find Cv.

In order to determine the heat flow from B to C, you have to determine the change in temperature using the Ideal gas law and determine the change in internal energy. The heat flow is the change in internal energy + work done (or you can use dQ = nCpdT since pressure is constant)

Repeat this for CD and DA (is the heat flow positive or negative?) and add them all together (careful with the signs) to get the total net heat flow.

AM

Last edited: Mar 25, 2008
6. Mar 25, 2008

### ~christina~

I still don't know how I'd determine that work was negative since the graph has the cylcle in a perfectlyshaped square so the volume both increases and decreases and my book doesn't explain this..

Making sure I understand you here:

For AB
There is no work done for when there is no change in volume so W= 0 and

$$dU= dQ - dW$$
W= 0 so
$$dU= dQ$$

and I know that $$\Delta U = nC_v\Delta T$$
and $$PV= nRT$$ and so $$\DeltaT= \Delta(PV)/nR$$

and since V doesn't change but P does then: $$\DeltaT= \Delta P V_i /nR$$

BC=> pressure constant
$$\Delta U=Q-W$$
$$Q= nC_p \Delta T$$
$$PV= nRT$$ and so $$\DeltaT= \Delta(PV)/nR$$

CD is the same as AB except

$$dU= dQ$$
$$\Delta U = nC_v\Delta T$$
and $$PV= nRT$$ and so $$\DeltaT= \Delta(PV)/nR$$

$$\Delta T= \Delta P 3V_i /nR$$

DA is same as BC except that the volume decreases...

$$Q= nC_p \Delta T$$
$$PV= nRT$$ and so $$\DeltaT= \Delta(PV)/nR$$

I assume that I substitute the pressure into the equations and then add them..but I'm not sure about the signs...how would I know if it si +/-?

7. Mar 26, 2008

### Andrew Mason

Work done ON the gas is negative if positive work is done BY the gas. The direction of the path determines whether work is positive or negative. Generally, work done BY the system is positive and work done on the system is negative. The area under the path BC is positive. The area under the path DA is negative. Add them together to get the net work.

So far so good. If the path is from left to right the area under the graph (work done BY gas) is positive.

AM

8. Mar 26, 2008

### ~christina~

how would I find a numerical value for part c if I don't know whether the gas is monotomic or diatomic? they give the temperature but..
and I thought I had to multiply to get the net work like I did in part A

okay, so this is what I got when I add them but I don't think I can get a number, correct?

just this=> $$nC_v \Delta(P V_i)/nR + nC_p \Delta(PV)/nR - nC_v \Delta (PV_i)- nCp \Delta(PV)/nR$$[

9. Mar 26, 2008

### Andrew Mason

The area under BC is 6PiVi and the area under DA is -2PiVi so the total is 4PiVi. This represents the net work done BY the gas. So the net work done ON the gas is -4PiVi.

$$C_v 2P_iV_i/R + C_p6P_iV_i/R - C_v6P_iV_i - C_p2P_iV_i/R = 4P_iV_i(C_p - C_v) /R$$

What is Cp-Cv for any gas?

AM

10. Mar 26, 2008

### ~christina~

Oh so you have to get the whole area below the box...I thought all I had to do was get the area in the box, that's why I couldn't figure it out.

but that would be their definition of a numercal value? You don't have to use the given temperature of 0 C ? and 1mol ?

Cp-Cv= R

so it becomes $$4P_iV_i$$

Last edited: Mar 26, 2008
11. Mar 26, 2008

### Andrew Mason

If you take the area under BC and subtract the area under DA the result is the area in the box - the net work done.

Use the ideal gas law to determine PiVi (n = 1, T = 273K). That will give you the numerical result.

AM

12. Mar 26, 2008

### ~christina~

okay I got
PV= nRT

PV= 1(8.31J/K*mol)*273K= 2432.43

4(PV)= 9729.72 J

Thank You

13. Mar 26, 2008

### Andrew Mason

That is the net work done by the system per cycle for part c).

Part b), the net heat energy added to the system is a not clear (the word "net" is confusing). If it refers to Qh - the heat flow into the system, it is asking for the heat flow in parts AB and BC only (heat flows out to the cold reservoir in CD and DA). In order to provide that as a numerical value you will need to know Cv.

AM

14. Mar 28, 2008

### ~christina~

so it would just be $$4P_iV_i$$ since after you cancel, with (Cp- Cv= R)

Thanks

15. Mar 30, 2008

### Andrew Mason

For A-C?

For AB the heat in is Cv(2Pi)Vi/R. For BC, Qh is Cp(2Vi)(3Pi)/R. So the total is:

$$Q_{hAC} = P_iV_i(2C_v + 6C_p)/R = P_iV_i(2C_v + 6(C_v +R))/R = P_iV_i(8C_v + 6)/R$$

AM

16. Mar 31, 2008

### ~christina~

oh..okay (I thought I was getting the total heat of one cycle in the above calculations but I guess I was calculating wrong.