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Need help with Thermodynamics question...


TokenMonkey

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Right, so I've got this Thermo assignment, and the question I'm having problems with looks like this:

 

Q3.gif

 

I can work out all the conditions in both compartments before the heating, using PV=nRT. There are two things I'm having problems with, however.

 

Firstly, to find the conditions AFTER the heating, I get 3 unknowns and 2 equations, which is obviously a problem.

 

What I have so far is this:

 

PA = PB, since I'm assuming energy was added to the system slowly enough such that we can take the process to be reversible.

 

Using the ideal gas equation for both compartments, we get

 

nATA/VA = nBTB/VB

 

and we also know that VA + VB = Vtot

 

The unknowns are VA, VB and TB.

 

Anyone have any ideas on where I can find a third equation?

 

Also, once I have worked out the conditions, I'll have to work out the amount of energy added by the heater. Since dU = 0 for an ideal gas, the energy balance reduces to Q = -W where W=-∫PdV. Am I right in saying I'll need to work out separate W terms for each of the two compartments? If so, how would I do this? My initial thought is simple enough, just use PV=nRT, but the problem is that the expansion of compartment A and the compression of B are not isothermal, so T cannot be taken as constant... How do I get around this?

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Since dU = 0 for an ideal gas

 

What makes you think this? The internal energy for an ideal gas is proportional to the product of the temperature and mass of the gas. The mass of the gas in each compartment is constant, but the temperature of these gases certainly is not constant, so neither is the internal energy.

 

What you need is a model of the thermodynamic process that the gas in compartment B undergoes. What does this tell you about that process?

If the cylinder and piston are heat insulators with negligible heat capacity ...

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Looks to me as if you need the 1st law of themodynamics.

 

http://en.wikipedia.org/wiki/First_law_of_thermodynamics

 

Specifically

 

dU = dQ - dW

dU = TdS - pdV

 

I take it you mean I need the First Law in order to work out the Q. Correct me if I'm wrong, but is what I wrote there not the First Law? What I mean is this:

 

You start with dU = dQ - dW

 

But since the gas in question is ideal, dU = 0 since ideal gases have constant internal energy (or is this a lie?).

 

So, 0 = dQ - dW, i.e., dQ = dW

 

Since W = ∫PdV, dW = PdV, so Q = ∫PdV

 

I think it's pretty much what I wrote, just not explained in-depth.

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What makes you think this? The internal energy for an ideal gas is proportional to the product of the temperature and mass of the gas. The mass of the gas in each compartment is constant, but the temperature of these gases certainly is not constant, so neither is the internal energy.

 

What you need is a model of the thermodynamic process that the gas in compartment B undergoes. What does this tell you about that process?

 

A-ha, so my belief is a lie! Umm... I could have sworn I read it in some notes somewhere, but chances are I remembered that statement and not the conditions which make the statement true! I thought it was a bit sketchy, but it was a convenient simplifier, not so? Hehe...

 

Anyway, that aside, what "If the cylinder and piston are heat insulators with negligible heat capacity ..." tells me is that compartment B is adiabatic, so the energy added by the heater will not directly affect it, only indirectly. The only work done on it is P-V work of its compression.

 

However, before I can even start thinking about that, I need to solve the first part, which is the conditions in compartment B... I still need to know one more variable... I need to work out either VA, VB or TB somehow. Is there any way I can relate the pressure in A before the heating to the pressure in A after the heating? That would be of immense help.

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You know the process is adiabatic and you assume its reversible (good assumption). Therefore there are only two unknowns because [math]\frac d{dt}(pV^{\gamma}) = 0[/math] for an adiabatic process, where [math]\gamma[/math] is the ratio of the isobaric and isochoric specific heats for the gas: [math]\gamma=\frac{c_p}{c_v}[/math]

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You know the process is adiabatic and you assume its reversible (good assumption). Therefore there are only two unknowns because [math]\frac d{dt}(pV^{\gamma}) = 0[/math] for an adiabatic process, where [math]\gamma[/math] is the ratio of the isobaric and isochoric specific heats for the gas: [math]\gamma=\frac{c_p}{c_v}[/math]

 

A-ha, so that (and Wikipedia) would lead me to believe that for an ideal gas, I can use [math]\gamma = \frac{5}{3}[/math]... Correct?

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I take it you mean I need the First Law in order to work out the Q. Correct me if I'm wrong, but is what I wrote there not the First Law? What I mean is this:

 

You start with dU = dQ - dW

 

But since the gas in question is ideal, dU = 0 since ideal gases have constant internal energy (or is this a lie?).

 

So, 0 = dQ - dW, i.e., dQ = dW

 

Since W = ∫PdV, dW = PdV, so Q = ∫PdV

 

I think it's pretty much what I wrote, just not explained in-depth.

 

I'm sorry I completely missed this.

 

Yes dU != 0. This is probably your problem.

 

I *think* you can do with without knowing gamma.

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My plan of attack was this: An energy balance on A will reveal dU + PdV = dQ. Integrating that whole lot and using dU = CvdT, I should get Q, not so?

 

HOWEVER, the issue arises in trying to calculate the PdV term in B (or A for that matter). W = ∫PdV, where P = nRT/V. Great, except for the fact that T isn't constant... How do I account for this in the integration?

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