How Does Ethane's Behavior Change Under Compression and Heat Transfer?

[erick]

1. Ethane, initially at 35 kPa with a volume of .12 m³, is
compressed without friction in a cylinder until its volume is halved
in such a manner that its pressure and volume are linearly related,
p = a + bV. The final pressure is 80 kPa and the change in internal
energy of the ethane is 3.22 kJ. Determine the heat transfer.

2. Ethane in a closed system is compressed without friction from 95
to 190 kPa in such a manner that pV = constant. Initially, the density
is 1.11 kg / m³ and the volume is 0.045 m³. During the compression,
2.96 kJ of heat is removed from the ethane. Determine the change in
internal energy of the ethane.

3. A closed vertical cylinder containing 0.3 kg of nitrogen at 90ºC is
fitted with a weighted, frictionless piston so that a constant pressure
of 275 kPa is maintained on the gas. The nitrogen is stirred by a
paddle wheel inserted through the cylinder wall until the absolute
temperature of the gas is doubled. During the process, 20 kJ of heat
is transferred from the nitrogen to the surroundings. Determine the
amount of paddle-wheel work required for this process.

 

[Saladsamurai]

We do not provide answers at PF, you are required to show some effort in order to receive help.

For# 1 I would start by writing out the energy balance for a closed system.

same for #2 and #3. We'll go from there.

 

[erick]

I'm having trouble solving... I've try a lot but still can't get it.
Could anyone write the solutions and with steps on how they did it so I can possibly learn! Thanks a lot

 

[Saladsamurai]

We do not provide answers at PF, you are required to show some effort in order to receive help.

When you created an account I am sure you read this. Post some work and we will post some help.

I will start you off for #1.

For a closed system, the energy balance reads: \Delta U=Q-W

You are given \Delta U and you need to find Q which means that all that is left is to calculate the Work done on the gas (i.e., it is being compressed, so work should be negative).

You are given initial pressure, final pressure and the fact that they are linearly related so you can find the work.

For work done in a piston-cylinder assembly, can you at least tell me how Work, pressure and Volume are related to each other in general?

 

[erick]

weee, I'm done with #1 but i don't know my answer if it's right. = -0.25 kJ

 

[Saladsamurai]

And neither do I since you still have not posted any work.

 

[erick]

Wnf = pV lnP1/P2
= (35 N / m²)(0.12 m³)[ln (35 kPa / 80 kPa
= -3.4692 x 10ˉ³ N-m
= -3.4692 kJ

Q = ΔU + W
= 3.22 kJ + (-3.47 kJ)
= - 0.25 kJ

 

[Saladsamurai]

Why, out of curiosity, do you think that that is the correct formula for the work done on the gas?

I asked you this relatively simple question in post #4 but, as I assumed it would, it went unanswered:

For work done in a piston-cylinder assembly, can you at least tell me how Work, pressure and Volume are related to each other in general?

 

[erick]

/sad. ok, I'm out. i can't solve it, but thnx for the help. goodluck..

 

[Saladsamurai]

So you do not know that W=\int P\, dV ?? You haven't seen that anywhere?

 

[erick]

hmmm, is that the correct formula?

 

[Saladsamurai]

What class is this for? You are trying to tell me that you have ever seen the expression Work=\int P\, dV?

 

[erick]

thermodynamics sir.

 

[Saladsamurai]

You are trying to tell me that you have ever seen the expression Work=\int P\, dV?

I feel like I have to ask you everything twice... Do you want help with this problem or not? Believe it or not I do have other things I could be doing... like nothing.

 

[erick]

i know that formula...

 

[Saladsamurai]

I give up. Clearly you don't want to make any effort towards solving this problem. Best of luck to you.

 

[erick]

ok..

 

[erick]

and tnx for nothing

 

[Saladsamurai]

Your are very welcome

Maybe if you took the time to participate instead of waiting for others to do your work for you, you would have enjoyed your time here better.

 

[erick]

W = ∫ P dV
= P ∫ dV
= p(V2 – V1)

what kind of effort do you want me to do? i'll try to solve it but still I'm wrong..

 

[Saladsamurai]

The problem is, is that you just keep guessing, which is why your answer is still wrong. I have been trying to get a little bit of a discussion going about the problem to see what you know and what you do not. A little courtesy is all we ask for here. When you signed up you had to agree to the terms and conditions, which explicitly say you must make an effort. If you want to review the rules, they can be found https://www.physicsforums.com/showthread.php?t=5374".

Now, the formula that you have W=p\Delta V is only good when you have constant pressure, which you do not.

So, in order to evaluate the integral, you need to write out pressure as a function of volume. You are given that

pressure and volume are linearly related, p = a + bV

and you are given initial and final conditions

thus you should be able write pressure as a function of volume, i.e. solve the two equations in two unknowns.

You know at p=35kPa, V=.12m³ and p=80kPa, V=.06m³ therefore:

35=a+b(.12)
80=a+b(.06)

so solve for the constants a and b, put them back into p=a+bV and you have your function p(V).