Energy change due to sudden expansion

[jasoncurious]

A gas is confined in a cylinder by a piston. The initial pressure of the gas is 7 bar(700kPa), and the volume is 0.10m^3. The piston is held in place by latches in the cylinder wall.The whole apparatus is placed in a total vacuum. What is the energy change of the apparatus if the restraining latches are removed so that the gas suddenly expands to double its initial volume, the piston striking other latches at the end of the process?

 

 

Data known:

P(initial) = 700kPa

V(initial) = 0.10m^3

V(final) = 0.2m^3

 

My analysis:

Since the gas suddenly expands (not a quasi-equilibrium process), the PV ln (V2/V1) equation cannot be used. But still, my lecturer gave me zero. I want to know the right way to solve this problem. God bless you all.

Edited November 24, 2012 by jasoncurious

[swansont]

What equation should you apply?

[jasoncurious]

If PV ln (V2/V1) were to be used, it doesn't make sense. The gas expands suddenly, not in a quasi-static manner. I don't know what other equation can be used.

[jasoncurious]

I was thinking if there will be no change in energy at all, i.e deltaE=0, since the exapnsion is done against vacuum?

[Axioms]

I am sure there will be a change in energy, acting on the walls of the system, because the volume increases and thus the pressure will decrease. If pressure decreases the energy of the atoms hitting the wall will be less because they have more space to move.

 

Try isolating the first part of the system. You can then look at the second part of the system in isolation as well. Once you have done this you can compare and compute the change in energy. Maybe this is what your lecturer was looking for? There is a little trick with the units that you can do. I will give you the start and you can then try figure out what I mean.

 

Remember Pa= N/m^2

J= Nm

 

To compare pressure and energy using the data that was given: J = Nm = (N/m^2)*(m^3) = (Pa)*(V); [(isolated values!)]

In words energy is equal to the pressure*volume of the system.

 

That is what I would have done. A good tip to answering questions like that is to understand what each equation actually means. Try breaking your equations down into parts and see how they were derived. Once you have that understanding all the questions regarding the equation become simple because most of the questions require you to solve parts of the equations simultaneously. If you know what you need from each equation it will make life simpler in solving the problem.

 

Edit: The internal systems energy will remain the same. But I am sure that is not what they were looking for because that is just theory. They would have also used the wording of internal system or something to indicate that they are looking for internal energy.

Edited November 26, 2012 by Axioms

[swansont]

I was thinking if there will be no change in energy at all, i.e deltaE=0, since the exapnsion is done against vacuum?

Right. It's not doing any work against a vacuum, and there is no change in the internal energy. PV = nRT and even though the system is not in equilibrium during the expansion, you know what happens at the end: the final T will be the same. Thus, PV is a constant.

[Axioms]

Right. It's not doing any work against a vacuum, and there is no change in the internal energy. PV = nRT and even though the system is not in equilibrium during the expansion, you know what happens at the end: the final T will be the same. Thus, PV is a constant.

 

I agree that the internal energy is constant.

 

If I do the calculation I wanted you to do:

 

Pressure(initial)*Volume(initial)= Pressure(final)* Volume(Final)

 

therefore: Pressure(final)= [Pressure(initial)*Volume(initial)] / [Volume(Final)]

therefore: Pressure(final)= (700kPa*0.1m^3) / (0.2m^3)

= 350kPa

 

Energy(initial) = 700kPa*0.1m^3

= 70kJ

 

Energy(final) = 350kPa*0.2m^3

= 70kJ

 

Thus there is no change in energy.

 

Haha sorry I was wrong in thinking that there was a change in energy but I did not do the calculation until now. I also never realised that they placed it in a vaccume. Im used to questions where atmospheric pressure changes so I guess I just went on that basis.

 

 

But still if you are not sure you can always just work it out. If it was for more than 1 mark then you would need a good explination, and showing the working will garantee your marks.

[alpha2cen]

This may seem like adiabatic expansion problem.

 

[latex]dW=-PdV[/latex]

 

[latex]P_{a}V_{a}^{\gamma }=P_{b}V_{b}^{\gamma } \quad ; \quad adiabatic \quad expansion[/latex]

 

[latex]\gamma \quad ; \quad heat \quad capacity \quad ratio[/latex]

 

More calculation is needed.

Edited November 26, 2012 by alpha2cen

[jasoncurious]

Thanks guys. At least I've sth to fight my case.