# Homework Help: Thermodynamics help.

1. Nov 28, 2008

### bubuchnik

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

1. An ideal gas with a molar mass of 40*kg/kmol and a specific heat ratio k*=*1.67 receives an amount of heat per unit mass equal to 10*kJ/kg. Calculate the resultant temperature rise if ,

a) The heat is added at constant volume
b) The heat is added at constant pressure

2. Hydrogen is contained in a piston-cylinder device at 100 kPa and 1 m3. At this state, a linear spring (F proportional to x) with a spring constant of 200 kN/m is touching the piston but exerts no force on it. The cross-sectional area of the piston is 0.8 m2. Heat is transferred to the hydrogen, causing it to expand until its volume doubles. Determine (a) the final pressure, (b) the total work done by the hydrogen and (c) the amount of work done against the spring. Also show the process on a p-V diagram.

If i knew the equations and the how to even attempt these i would post them. But i'm truly stumped !! :(

Any help would be appreciated. Thank you...

EDIT: If you really need to see some work before you can help...

1)For the first one i tried applying q=mc(delta)t
And since c for constant volume is 12.41 (k= cv/cp)
I'm getting (delta)t as around 20. But the answer says 32.3
And following the same process for constant pressure i'm gettin about 12 but the answer says 19.3

2) I tried to find the final pressure first. And i did p1/p2=v1/v2 and since the volume doubles p2 should be 50 kpa. But the answer says 412.5 kpa. Since the rest of it depends on this i didn't try further.

I know i'm missing something big here.

Last edited: Nov 28, 2008
2. Nov 28, 2008

### Andrew Mason

Using the first law of thermodynamics: dQ = dU + dW. For constant volume, dQ = dU (since dW = PdV = 0). Since dQ = nCvdT you just have to find Cv. So dT = dQ/nCv. So far, so good. Since it is a monatomic gas, Cv = 3R/2 = 12.47 J/K mol. For constant pressure, you use Cp = 1.67Cv.

It is your determination of dQ that is the problem. The molecular weight of the gas is needed to convert the heat/kg to heat/mole. Let's assume one mole of gas. How much heat is added?

You cannot use p1/p2=v2/v1 unless the temperature is constant. Use the first law: dQ = dU + PdV. Since the spring force changes with distance, the pressure is a function of volume. Use that relationship to determine the work done (PdV).

AM

Last edited: Nov 28, 2008
3. Nov 29, 2008

### bubuchnik

Hey,

I figured out the first question, thanks :D

The second one i'm confused. How do i equate pressure to volume ? Do you mean PV=mRT ?

But then what are dq and qu ?

I'm confused...

4. Nov 29, 2008

### Andrew Mason

The pressure in the cylinder is the force on the piston / area of piston. What will provide this force? Hint: they give you the spring constant for a reason.

AM

5. Nov 30, 2008

### bubuchnik

Hey just want to clarify. For the first question i'm getting the answer if for cp-cv=R i use the universal R value of 8.314. Is that right ?

And for the second one i wrote if the area is 1m^3 and area of piston was .8m^2 then the length would be 1.25 m. For volume to be 2m^3 the length would be 2.5. So it would increase by 1.25m and thus the pressure would be initial pressure + (k*1.25)/.8, which give me the correct answer.

And for the work done by hydrogen. What would be dq, dw and du? I don't get this....

6. Nov 30, 2008

### Andrew Mason

This is correct but unless you show your work you will have problems with the rest of it. Set out the expression for Pressure as a function of volume.

To determine the work done by the hydrogen, you have to calculate PdV over the whole expansion (Hint: you need to do an integration).

Can you work that integral out using the expression for P ?

AM

7. Dec 3, 2008

### bubuchnik

K so i related volume to pressure.

My extension was volume/area (v/a)
And force was 200 * (v/a)

But since the spring didn't act till length was greater than 1.25

f=(v/a - 1.25) *200

And pressure was f/a. And since initial pressure was 100kpa i added that to the p. So p= f/a + 100
Which simplified to p = 312.5v - 212.5

Correct ? I'm getting the answer for final pressure correct, it's 412.5 kpa.

For work done by hydrogen i integrated the equation between 1 and 2 and got the answer. It was 256.25 kj

But it asks what's the work done against the spring. I'm confused here now. Help ?

8. Dec 4, 2008

### Andrew Mason

Correct, but not clear. You are not showing your work. You should use variables to do the analysis and then plug in the numbers: eg

$$P = P_0 + F_{spring}/A_{Piston} = P_0 + kx/A = P_0 + \frac{k\frac{V}{A}}{A} = P_0 + \frac{kV}{A^2}$$

Work done is:

$$W = \int_{V_i}^{V_f} PdV = \int_{V_i}^{V_f} \left{(}P_0 + \frac{kV}{A^2}\right{)}dV = \int_{V_i}^{V_f} P_0dv + \int_{V_i}^{V_f}\frac{kV}{A^2}dV$$

As seen from the above expression for Work (W), the total work is the sum of the work done against the atmospheric pressure and the work done against the spring. You can work that out.

AM

9. Dec 4, 2008

### bubuchnik

I've handed it in. Yes i did write it using variables and properly. Just that it's hard doing that on an internet with proper fractions and stuff. So i was just posting my logic.

I actually said the total work was p.dv and since the final pressure was 412.5kpa and the change in volume was 1m3 the total work was 412.5 kj. So wor done by piston was 412.5 - work done by hydrogen. Got me the correct answers...hope my logic was right.

10. Dec 4, 2008

### Andrew Mason

These answers are not correct and the reasoning is not correct.

The final pressure is 412.5 KPa. The final force from the spring is 250 kN and final pressure from the spring is 312.5 KPa/m^2, but the piston does not work against that pressure throughout the whole distance. The correct answer is:

Total Work:

$$W = \int_{V_i}^{V_f} P_0dv + \int_{V_i}^{V_f}\frac{kV}{A^2}dV$$ where V is the expansion (ie Vi = 0 and Vf = 1)

$$W = P_0V + \frac{k(V^2)}{2A^2}$$ where V = 1 m^3

$$W = 100(1) + 200*(1)/2(.8*.8) = 256.25 KJ$$

So the total work done is 256.25 KJ and the work done against the spring is 156.25 KJ.

AM