Calculating Energy Lost in Gas Expansion

[aaaa202]

Suppose I have a gas expanding. The gas is an isolated system and has an internal pressure. Outside there is another pressure.

If I then want to find how much energy the gas loses, do I then see it as the internal pressure of the gas doing a positive work on the atmosphere or the atmospheres pressure doing a negative work the gas?

Because the two pressures are different and I don't see which pressure is better using - both procedures make sense to me - which one is wrong and why?

 

[ehild]

Think of the First Law: The Internal energy of the gas changes by adding heat and/or doing work on the gas. If the internal pressure differs from the outer one the gas is not in equilibrium during the expansion, its pressure is not defined.
The work done by the external agent is of the same magnitude as the work of the gas if the process is quasi-static, so slow, that the gas is practically in equilibrium during the whole process. Such process is also reversible.
Free extension of the gas is not a reversible process.

ehild

 

[aaaa202]

so adiabatic compression and all those processes are only valid when the a gas expands with internal pressure equal to external?

 

[aaaa202]

I just think it's weird that you can assume that. If the gas is in equilibrium it won't move either way, so there must a net force on it?

 

[ehild]

It is almost in equilibrium, the internal pressure is just a bit over the external. The net force can be as small as you like, still it will cause expansion. Our teacher explained that on a gas in a vertical cylinder fitted with a piston, and some sand on the piston, to keep equilibrium. He said that a quasi-static process means blowing off a grain of sand, waiting till the piston moves to equilibrium position again and then blowing off the next grain.

ehild

 

[ehild]

so adiabatic compression and all those processes are only valid when the a gas expands with internal pressure equal to external?

You can use the work of the gas expressed with the pressure if the pressure is defined at all. If the gas is not in equilibrium, the intensive parameters, pressure, temperature are not defined.

ehild

 

[aaaa202]

Hmm okay it makes sense, but then consider this exercise:

"suppose you have a container of helium. This container is somehow made to expand such that pressure and volume are directly proportional."

Would this be possible? And how could you do it? Elastic walls?

 

[ehild]

You can change the volume by a piston. The volume increases if the pressure of the gas exceeds the external force on the piston. You can make the pressure increase by rising the temperature. So warming up the gas both the volume and the pressure can increase.ehild

 

[aaaa202]

hmm... Yes p is proportional to T if V is constance. Also V is proportional to T if p is constant. But are you sure that you can just say that both are proportional to T if both of them are changing?

 

[ehild]

hmm... Yes p is proportional to T if V is constance. Also V is proportional to T if p is constant. But are you sure that you can just say that both are proportional to T if both of them are changing?

No, why should they? The pressure and volume can both change in different ways. For example, they can change such way that that the temperature stays constant. That is an isotherm process.

The question was

"suppose you have a container of helium. This container is somehow made to expand such that pressure and volume are directly proportional."

Pressure and temperature are proportional to each other. That does not mean that they are both proportional to the temperature.

Imagine you have 1 m³ helium at 100 kPa pressure and 300 K temperature. At what temperature becomes both the pressure and temperature twice the initial value?

ehild

 

[aaaa202]

Wuuut? p and V are inversely proportional if the gas is to fulfill the ideal gas law.

 

[ehild]

I am sorry, I can not do more for you. Read your lecture notes about ideal gas law.

ehild

 

[aaaa202]

sorry yes I got it. You suggest that we increase the volume whilst also making sure than the temperature increases sufficiently such that p and V are proportional.