Calculating work done on parcel given a specific process?

[Foxhound013]

Homework Statement

Problem: A sample of dry air has an initial pressure p₁ = 1000 hPa and temperature T₁ = 300K. It undergoes a process that takes it to a new pressure p₂ = 500 hPa with unchanged temperature T₂ = T₁. Compute the mechanical work per unit mass perfromed by the sample under the following scenarios.
a) Isochoric pressure reduction to p₂ followed by isobaric expansion to final state.
b) Isobaric expansion to final specific volume α₂ followed by isochoric pressure reduction to final state.
c) Isothermal expansion to final state.

A pressure by specific volume graph may be helpful.

In a nutshell the given data is as follows
T₁ = T₂ = 300 K
p₁ = 1000 hPa
p₂ = 500 hPa

-Gas constant R_d = 287 JK^-1kg_-1

Homework Equations

pressure volume work equation: w = p*dα
equation of state: pα=R_dT

The Attempt at a Solution

Part a) I started by breaking down the process listed in part a. I drew a pressure by specific volume graph starting first with moving the parcel under constant volume conditions to the new pressure and then moving it to the new specific volume under constant pressure.

After this I looked at each piece as a leg in the journey of this air parcel and started figuring out the work involved in the first leg.

∫ dw = ∫ pdα
w = 0 I arrived at this solution for the first leg because the parcel doesn't change in specific volume (due to it being isochoric) meaning that dα = 0 . . . so w = 0.

The second leg involved changing the parcels specific volume under constant pressure. I applied the same idea a second time.

∫ dw = ∫ pdα The limits of integration are from α₁ up to α₂.
w = p(α₂ - α₁) From this point I rearranged the equation of state and solved for α
α = (R_dT)/ p
So. . .
w = p₂ ( ((R_dT)/ p₂)) - ((R_dT)/ p₁) )

plugging in the values w works out to be 43050 Jkg^-1.

Speaking with my professor today, he says that the work done should be negative? Why should it be negative?
Since the parcel is expanding it is doing work on the environment, therefore the work is positive right?

What am I missing?

I calculated part B in almost exactly the same fashion but arrived at a -43050 Jkg^-1.

As for part C, the process I used was slightly different.

∫ dw = ∫ pdα Using the equation of state I substituted for p.
∫ dw = ∫ (R_dT / α) dα
∫ dw = R_dT ∫ 1/α dα Limits being α₁ up to α₂
w = R_dT ( ln(α₂ / α₁))

I again used the equation of state to substitute for each of the α values

w = R_dT ( ln( (R_dT/ p₂) * (p₁ / R_dT) ) )
This simplifies down to . . .
w = R_dT ( ln(p₁/p₂) )
Plugging in the corresponding values
w = 287 * 300 * (.693147181)
w = 59679.97 Jkg^-1Why is this answer different than the first two? Is it because of the process involved or because of a calculation mistake on my part?Thank you for any help you can provide, please let me know if I wasn't clear enough on any of the above parts and I'll try my best to clarify.

 

[Chestermiller]

My interpretation of the problem statement is the same as yours, and my results are the same as yours, except for part b. Please show your work.

Regarding the sign of the work, it is positive (unless, in your course, they use the sign convention that W is positive if it is work done on the system and negative if it is work done by the system).

Chet

 

[Foxhound013]

Here is my work for part b,

The first leg in this parcels journey is expansion under constant pressure to α₂.
So . . .
∫ dw = ∫ pdα with the limits of integration being α₁ up to α₂.
w = p (α₂ - α₁) I used the equation of state to substitute for α. R_d is written simply as R here
w = p₁ ( RT/p₁ - RT/p₁) Since the pressure level doesn't change p will always be the same in this portion.
w = 0

Proceeding to the second leg of this parcels journey, the parcel will undergo a reduction in pressure at a constant volume.
∫ dw = ∫ pdα with the limits of integration being α₁ up to α₁.
w = p (α₁ - α₁) I now use the equation of state to substitute for α since it is at two different pressure levels. If something is wrong, I get the feeling that this step will be it. I'm not certain how to deal with this though.
w = p ( RT/p₂ - RT/p₁)
w = RTp₁ / p₂ - RT
w = -43050 Jkg^-1

I believe the convention we're to use is work done by the system is positive and work done by the surroundings on the system is negative. My professor may have made a mistake since he didn't really look at the problem too closely when I asked him.

Thank you for the help! I really appreciate it!

 

[Chestermiller]

Here is my work for part b,

The first leg in this parcels journey is expansion under constant pressure to α₂.
So . . .
∫ dw = ∫ pdα with the limits of integration being α₁ up to α₂.
w = p (α₂ - α₁) I used the equation of state to substitute for α. R_d is written simply as R here
w = p₁ ( RT/p₁ - RT/p₁) Since the pressure level doesn't change p will always be the same in this portion.
w = 0

This assumes that the temperature at the end of step 1 is the same as the original temperature. But, to cause the volume to increase isobarically, you need to add heat to the system and raise the temperature. In the second step, you then hold the volume constant, and remove heat to lower the pressure isochorically (and lower the temperature to its original value). In step 1, you are holding the pressure constant, and increasing the volume. So, how much work is done?

 

[Foxhound013]

I think I'm following, but if I change the temperature in the first step, how can I go about calculating the work since I don't know anything about temperature?

Given what you've said for the first step w does not = 0. Is that correct?

 

[Chestermiller]

I think I'm following, but if I change the temperature in the first step, how can I go about calculating the work since I don't know anything about temperature?

Given what you've said for the first step w does not = 0. Is that correct?

Yes, the work in the first step is not zero. It doesn't matter what the temperature is doing if you know how the pressure and volume are varying. In part b, step 1, you know that the pressure is 1000 and you know the initial and final volumes of the gas (because you know that step 2 is carried out at constant volume, and you know the final volume at the end of step 2). You also know that the work is equal to the integral of PdV.

Chet