Recent content by hyddro

Thank you! That ws really helpful!

ok, I will do that tomorrow when I get back home... what if my sheet here at school is asking me to calculate the work... like a numerical value? Do i just calculate one side of the equation?

Oh God, i keep forgetting to address your question. Yes, they were absolute pressures.

ok, so I got it down to ΔlnT/ΔlnP = 1 - 1/γ what do i do next? I am still puzzled as to how this is going to help me find the work done by the expanding gas... :/ thanks though

so then, R ΔlnP = ΔlnT (Cv + R) ?

ΔU = Cv * ΔlnT = R * Δ(lnV) = R (ΔlnT - ΔlnP) do I keep going? Cv ΔlnT = R*ΔlnT - R*ΔlnP -> ΔlnT ( Cv - R) = - R*ΔlnP ? still lost... sorry :( (could you share a link where I can read about this topic, maybe i can save you some time if you could just direct me to the right topic)

Thank you. That is indeed very difficult to visualize... i can see the difference between this and piston that just allows the gas to expand rather than expand and escape. integrating gives me ΔU = nCv*ΔT = nRTln(v2/v1) is this correct? then what? do i use the left side of the...

I am having difficulties trying to understand the external pressure... You are saying that the pressure of the gas 'escaping' is now the external pressure of the gas that remains inside? How then do I calculate the work done by this expansion? w = nRT ln(v2/v1) ? but T isn't constant.. I am...

Also, we did not intend to reach any specific final or initial pressure reading (due to the difficulty of the experiment itself, holding the stopper of the flask was very difficult to do and thus we needed to be quick). Due to this, our readings vary a lot and thus the degree of expansion is...

well, i know the temperature change because the probe was measuring the temperature inside of the vessel at all times... what is the difference between a reversible and irreversible expansion... sorry but i took p.chem long ago and i forgot already some stuff... EDIT: Also, in a reversible...

Absolute pressure. The gas is allowed to expand against the atmosphere, no valve involved. Because the stopper was opened for a split second and thus some of the compressed air pushed against atmosphere (left the vessel) while still some remained slightly compressed. I think I didnt explain it...

isnt the gas pushing against the atmosphere?

Sorry. The vessel had a stopper that you could open and let the gas expand... That's what i meant..

I have come up with something. Since w = p (V2 - V1) i am thinking... w = p ( nRT2/P2 - nRT1/P1) ... assuming n = approximately constant... would this work? thanks

Homework Statement Hi, I did a lab experiment where I took a 5L vessel made of some material that isolates gas inside and thus behaves like an isolated system (adiabatic). I then pumped gas from ~90kPa to up to 150 kPa... recorded the temperature, then let the gas 'expand' by opening the...