As it can be read here, http://en.wikipedia.org/wiki/Laplace_transform#Relation_to_power_series
the Laplace transform is a continuous analog of a power series in which the discrete parameter n is replaced by the continuous parameter t, and x is replaced by exp(-s).
Therefore, computing a...
I imagine that it can be explained by the fact that the balloon "walls" are flexible and then equilibrate both pressures. Therefore, when the balloon is inside and outside the fridge, it actually has the same pressure, which is the atmosphere pressure.
Is this right?
If a helium-filled balloon initially at room temperature is placed in a freezer, will its volume increase, decrease, or remain the same?
The Attempt at a Solution
If you decrease the T of a vessel filled with gas, its molecules will have less kinetic energy, and the pressure inside the...
OK I think I am understanding it better, thanks! I have on more question. The surface calculated with an integral is usually not a rectangle such that area = length*height. However, I suppose that what you can do is "rearrange" the surface keeping the area constant and making it look like a...
Homework Statement
I have a question that looks so stupid that I have never dared to ask.
If I want to measure the time average from t=0s to t=1s of a given f(t), the solution is compute the following integral:
TA = 1/T*∫F(t)dt
However, I have some doubts about this calculus.Homework...
So how would you do to get a more accurate result? One option is to get a (T,p) point closer to the desired 69ºC and 1bar, but if this is not available...?
Homework Statement
The normal boiling point of hexane is 69.0°C. Estimate (a) its enthalpy
of vaporization
The Attempt at a Solution
In order to solve for ΔHvap in the Clausius-Clapeyron equation I need a reference T and p point. I extracted it from the triple point (T=178K and p=1,23Pa)...
Hello,
Wikipedia states: Phase changes, such as melting or evaporation, are also isothermal processes.
I am interested in calculating the enthalpy of a given phase transition.
If the process is isothermal, I would immediately say that H is 0, according to the following equation:
dH=CpdT...
1) What is the reason why dH!=0 for an adiabatic(q=0) reversible process?
The mathematical argument is irrefutable and it is clear that it has to do with the process not being isobaric:
ΔH=ΔU+PΔV+VΔP , ΔU=work=−PΔV
Therefore, ΔH=VΔP and this is not 0.
However, I do not understand it...
Ok thanks!
Concerning the third equation I forgot to say that it is in a context of a diabatic expansion (q=0). Therefore, ΔU=CvΔT=W.
Then in a diabatic expansion, we can have work only dependent on the T and not on the V. How does this not contradict the "classical" definition of W=pΔV?
Hello,
I have been self-learning Thermodynamics and I am having a bit of trouble with calculating the work in different circumstances.
Along the lectures we have come up with three different equations for work
1) W = pΔV
2) W = nRTln(V2/V1)
3) W = CvΔT
So my questions are:
1) which...