Pressure or volume for state thermodynamics

[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

 

[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.

 

[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.

 

[JSBeckton]

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.

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

 

[Cyrus]

You are given:$Gas=CO_2$, $P_1 =300kPA$, $T_1 = 100C^o$, [tex]V_1 = 0.2m^3 [/itex], and $T_2 = 200 C^o$.

From the initial state you know that:

$$V=vm$$

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.

 

[JSBeckton]

Thanks, I don't 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.

 

[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.

 

[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)

 

[Cyrus]

Nice. Did not see that. That is a lot 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.

 

[Nomy-the wanderer]

How about: W= R (T2-T1)/n-1

Am i too late??

 

[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?

 

[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.