# Pressure or volume for state thermodynamics

1. Mar 19, 2006

### JSBeckton

Dont know how to start this one without a pressure or volume for state 2-

A piston cylinder arrangement (cylinder facing up with a piston pushing down) contains C02 at 300kpa and 100 deg C with vol of .2m^3. Weights are added such that the gas compresses according to PV^1.2= constant to final temp of 200 deg C. Determine work done.

This problem was in the book but not assigned, we didn't do anything like it but I wouldn't put it past him to put it on the test anyways. Can anyone point me in the right direction?

Anwser -80.4 kJ

2. Mar 19, 2006

### Cyrus

This is a polytropic process.

$$P=CV^{-n}$$

$$W_b = \int^2_1 PdV = \int^2_1 CV^{-n} dV = C \frac{ V_2^{-n+1} - V_1^{-n+1}}{-n +1} = \frac{P_2V_2 - P_1V_1}{1-n}$$

Try and work it out now.

Last edited: Mar 19, 2006
3. Mar 20, 2006

### quark

Suggestion: Can the universal gas equation be of use to get the final state pressure volume product(i.e P2V2).

Note: Be sure that you are using absolute units.

4. Mar 20, 2006

### JSBeckton

Thats what I was trying to do but can't solve it because I dont have P2 or V2, if I had one I could solve for the other.

5. Mar 20, 2006

### Cyrus

You are given:$Gas=CO_2$, $P_1 =300kPA$, $T_1 = 100C^o$, $$V_1 = 0.2m^3 [/itex], and $T_2 = 200 C^o$. From the initial state you know that: [tex] V=vm$$

This means the initial mass, which is your total closed system, is equal to: $$\frac{V}{v} = m$$

You can find $v_{1,100C^o}$ from your tables.

You can also find $v_{2,200C^o}$ from your tables.

From there you can find $$V_2 = v_2 m$$

Now you can use:

$$PV^n = C$$

to find the pressure at state two.

Edit: If you don't want to use your tables you can also use EES, which will do all this lookup work and calculations for you. It's a very nice and very powerful equation solver.

Last edited: Mar 20, 2006
6. Mar 20, 2006

### JSBeckton

Thanks, I dont have tables for CO2 so they probabally wanted me to use the computer program that was with the book. Thankfully that question was not on the test.

7. Mar 20, 2006

### Cyrus

Are you using Cengel and Boles? Does it have a picture of the sunset on it? Its a VERY good book. I would tell you to go buy it if you have the cash and throw away your currenet book.

Last edited: Mar 20, 2006
8. Mar 20, 2006

### quark

You can do it without tables by using universal gas equation.

P1V1/T1 = P2V2/T2 or P2V2 = P1V1T2/T1

So P2V2 = 300 kPa x 0.2 cu.m x 473 K/373 K (based upon the answer 300 kPa seems to be absolute pressure).

So, P2V2 = 76.08 kJ

So work done is (76.08 - 60)/(-0.2) = -80.4 kJ (negative sign indicates work done on the system)

9. Mar 20, 2006

### Cyrus

Nice. Did not see that. That is alot easier.

BAH! I am looking at my tables now, I forgot that you need two properties to specify the specific volume at a state. So knowing just the temperature is not enough. The Pv=RT relation is the ONLY way to solve it.

Last edited: Mar 21, 2006
10. Mar 22, 2006

### Nomy-the wanderer

Am i too late??

11. May 16, 2006

### jasc15

Is polytropic the same as adibatic? (PV^n)
Edit: i think i have the answer: it is adibatic if n=cp/cv, isobaric if n=0 and isothermal if n=1. But that raises another question of course, what if 0<=n<=1?

Last edited: May 16, 2006
12. Aug 8, 2006

### Djomloun

All process are polytropic, but there are same special casess for n=0, n=1 etc so I thing that for 0<n<1 it is a polytropic process.