# Paypal/nochex waiting as a thankyou

Two gas cylinders A and B for which Va = 30 litres and Vb = 10 litres are linked by a narrow pipe containing a valve. Initially the valve is closed and cylinder A is charged with an idea monotomic gas to a pressure Pa = 10Ma at Ta = 300K. Vessel V is evactuated and has pressure Pb = 0. The valve is slowly opened and the gas pressures allowed to equalise and the final temperature Ta = Tb = 300k.

a) Is the process quasistatic or not? Give reasons.
b) What is the final pressure? What is the change in internal energy?
c) Calculate the external work performed on the system.
d) Determine the total heat transferred to the system.
e) Would the answers to a-d change if the process was a free expansion rather than a controlled leak?

Many thanks

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Hmmmm....the question I originally replied to seems to have disappeared...strange

[For the first bit, I would assume that you need to start from another adiabatic relation (something involving pressure and volume perhaps...) and then use the ideal gas equation pV = nRT to relate that to what you want... (remember that n and R are constants).

For the second part, call the initial temperature and radius/volume T1 and V1, and call the final temperature and radius/volume T2 and V2. Since

T1.V1^(lambda - 1) = A

(A is a constant)

and

T2.V2^(lambda - 1) = A as well

you should be able to rearrange these equations to get V2, and hence the final radius.

Hope that helps...

Jess]

This is all a bit pointless now...sorry

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I was just in the process of changing the question!
Thanks for you help, I think I have that one sussed now though cheers.

Any chance you can have a go at this one?

Rob.

Are you sure Pa = 0? That seems a bit odd to me, but I could be reading the question wrong.

Jess

Thanks again Jess! lol I think I'm having a bad day!

I'm guessing the answer to a) would be that is is a quasistatic process due to the pressures being allowed to equalise. But are there any better reasons for this?

Rob

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Originally posted by robgb

I'm guessing the answer to a) would be that is is a quasistatic process due to the pressures being allowed to equalise. But are there any better reasons for this?

Rob

I would think that that answer would do, although I think the part about the pressures equalising needs to be more specific - you might want to define exactly what you think a quasistatic process involves to let whoever marks this (I assume someone will, at some point) know that you know what you're talking about.

You might also want to think about what part (e) (free expansion) involves and how this is different from a quasistatic process - this should make the answer to (a) a lot more obvious.

As a clue for part (b) (the internal energy bit), since we are talking about an ideal or perfect gas, I suggest you look up Joule's law...

I don't know what level you're at, but I've always found a book by C.J. Adkins called 'Equilibrium Thermodynamics' to be a good source of information for this sort of stuff.

Jess

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Thanks Jess.
Could you shed any light onto c & d?
Any chance anyone came come up with an answer?
These are questions my son have been set, and I'm trying to get some answers so that I can help him with them.

Rob.