6
votes
Why expansion of real gases lead to cooling?
There are various conditions which might apply while a gas expands.
adiabatic (isentropic) expansion. Gas does work on its surroundings (pushing on the boundary as the boundary moves) so loses energy....
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3
votes
Why expansion of real gases lead to cooling?
The reason given in the book stays on a kinetic theory level of description. I think it is not the simplest level to explain the phenomenon. It is much better to start acknowledging that in a free ...
2
votes
Accepted
Find the ideal gas law from the internal energy
I am not totally sure if this is the most straightforward way, but you will receive the ideal gas law by varying $U$, that is
$$\operatorname{d}U = \dfrac{\partial U}{\partial V} \operatorname{d}V + \...
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2
votes
Why is $\delta U$ not zero in $\delta H = \delta U + \delta(PV)$ while defining at a constant temperature for an ideal gas?
Your reference is describing an ideal gas mixture, not a single molecular species, which is undergoing a chemical reaction (presumably at constant temperature). For a gas mixture, the internal energy ...
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2
votes
Why is $\delta U$ not zero in $\delta H = \delta U + \delta(PV)$ while defining at a constant temperature for an ideal gas?
The internal energy of an ideal gas with $f$ degrees of freedom is $U = \frac{f}{2}nRT$. Under the assumption that $n$ is constant we can say that $U = U(T)$. However in your problem the number of ...
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1
vote
Why expansion of real gases lead to cooling?
Let us consider gas molecules collision with the piston to be fairly elastic in nature. While we expand the gas by lifting the piston with some velocity u. And let the gas molecules be traveling ...
- 117
1
vote
What is the average distance $\langle r \rangle$ of a particle in an ideal gas in a potential from the origin?
...The Hamiltonian is
$$
H(q^N, p^N) = \sum_{i=1}^{N} \frac{\textbf{p}^2_i}{2 m} + \sum_{i=1}^N u(\textbf{q}_i).
$$
The particles are not interacting with each other and the system is in contact with ...
- 17.6k
1
vote
What's happening in this approximation for the equation of state of an ideal gas of bosons
A few things have dawned on me, enough to make me feel like I understand the answer to my own question, with the help of the comments. So to any that may need it, here is what I have concluded.
First ...
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1
vote
Mixing entropy for ideal gases is decreasing
You are not comparing the same things. The entropy of the Sackur-Tetrode equation is:
$$
S = \ln\frac{\Gamma}{N!}
$$
while your subadditivity argument uses:
$$
S = \ln\Gamma
$$
Physically, your final ...
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