Thermodynamics: heat transfer

[WTST]

Hi,
I'm having some problems with this exercise on thermodynamics:
"A cylinder with a movable piston contains a certain quantity of helium. With a very slow transformation - represented in a V-p graph as a straight line - it goes from state [latex]A(p_A=40kPa, V_A=3dm^3, T_A=300K)[/latex] to state [latex]B(p_B=150kPa, V_B=1dm^3)[/latex]. It is then subjected to a costant-volume transformation followed by a costant-pressure one, returning to the initial state".

It can be summarized in this graph:
all
From which it's pretty simple to find the work done (area of the triangle):
[latex](110kPa*2dm^3)/2=(110Pa*2m^3)/2 \Rightarrow \color{red}{W=-110J}[/latex].
A clock-wise transformation means positive work, but this graph - misdirected by the text of the exercise - is all the opposite (V-p, instead of p-V), hence the negative work.

It then asks for [latex]T_B[/latex], which I calculated from the ideal gas law. I didn't know the amount of substance, but I took that from state A.
[latex]n=\frac{p_AV_A}{RT_A} \Rightarrow \color{red}{n=0,048}[/latex] then [latex]T_B=\frac{p_BV_B}{nR} \Rightarrow \color{red}{T_B \approx 375K}[/latex].

Finally, it wants the heat transfer with the outside during [latex]A\rightarrow B[/latex].
At first, I wanted to use [latex]Q=cm\Delta T[/latex], but the specific heat gave me some doubts: which one should I use when the transformation is not particular (I'm referring to the fact that there are specifc specific heats when the transformation is with costant pressure/costant volume)?

Then I thought I could use [latex]\Delta U=Q-W[/latex]. I could calculate the difference in internal energy:

[latex]\Delta U=\frac{3}{2}NK_b \Delta T \Rightarrow \Delta U=\frac{3}{2}(n*N_A)K_b \Delta T \Rightarrow \Delta U=\frac{3}{2}*2,89^{22}*1,38^{-23}*75 \Rightarrow \color{red}{\Delta U_{A\rightarrow B} \approx +45J}[/latex].

 

Since it is work done on the gas, it is more internal energy, not less: [latex]\Delta U=45=Q \color{red}{+} W[/latex]. Also, we are only talking about the work done from A to B, which includes the area underlying the triangle: [latex]\color{red}{W_{A\rightarrow B}=220J}[/latex]. But I can't figure out the sign of the Q: the work causes a compression, but I don't know how much. I mean: it could be so much that the temperature went much higher than 375K, arriving then at that level giving off heat; but it could also be that the work brought the temperature under 375K and that it arrived at that level absorbing heat.

And anyway, whatever sign I use the answer [latex](Q=-145J[/latex], the book says) isn't right.

 

So, where's my mistake?

Thanks in advance.

(If something isn't clear, please tell me)

Edited August 25, 2014 by WTST

[studiot]

Unfortunately I have been away for a couple of weeks and have just come across this one.

 

However we can work through it if you are still interested, there is much insight to be gained from this one.

 

I will call the third vertex of the triangle C.

 

 

So, where's my mistake?

 

Well I agreed with your calculations in parts (1) 110J and (2) 375K ,

and I agree that if you calculate the change in internal energy from A to B and subtract the work (with due regard to sign) you will obtain the heat change

but I calculated the work from A to B to be 190J, not 220J as you have it.

 

Edit

 

Also, we are only talking about the work done from A to B, which includes the area underlying the triangle:

I would guess that you have added the area between the BC and the pressure axis (110J) to the triangle. This is of course the leg where no work is done since the volume change is zero.

You should be calculating the area between AC and the Volume axis (80J), which will add to get you the proper answer.

 

You correctly calculated the change in internal energy to be 45J so (45 - 190)J = -145J, the heat change required.

 

 

You have correctly identified Cv as being 3R/2, but did not specifically note that this is because helium is a monatomic gas (you need to use all the facts given in the problem).

 

This makes Cp = 5R/2 since we are using ideal gas laws as the equations of state.

 

Why do you think it was important for the questions to state that the change AB was made slowly?

 

 

(I'm referring to the fact that there are specifc specific heats when the transformation is with costant pressure/costant volume)?

 

When we have constant volume conditions we use Q = CvdT; This can therefore be done for the leg CA of the cycle.

 

When we have constant pressure conditions we use Q = CpdT; This can therefore be done for the leg BC of the cycle.

 

However in general when the conditions are neither constant volume nor constant pressure we cannot directly use the specific heats but must use the equations

 

for internal energy or enthalpy.

 

http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/shegas.html

Edited August 30, 2014 by studiot

[studiot]

I see there have been over 40 new views since I posted my belated response, but no one else has shown enough interest to post as well.

 

This is a pity since we may have lost/driven away a potential member actually interested in technical matters, and also because as I said this is a nice question that demonstrates many useful ideas if one follows the the state and exchange variables around the cycle, by direct calculation rather than using the area of the triangle.

 

All the numbers work out nicely and the integration to find the work for the leg AB is easy but very instructive since you will not find it as a standard method in any textbook.

 

Any takers?