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ICE Princess25 [194]
4 years ago
11

The internal energy of a system is always increased by __________. a adding heat to the system and having the system do work on

the surroundings b having the system do work on the surroundings c withdrawing heat from the system d a volume compression e adding heat to the system
Physics
1 answer:
Nitella [24]4 years ago
8 0
There are two correct options:
<span>d) a volume compression
e) adding heat to the system

In fact, we can have a look at the first law of thermodynamics:
</span>\Delta U = Q-W
<span>where
</span>\Delta U is the variation of internal energy
<span>Q is the heat absorbed by the system
W is the work done by the system on the surrounding

It is important to keep in mind the right sign convention for Q and W: Q is positive when the heat is absorbed by the system, while it is negative when it is released by the system; W is positive when it is done by the system on the surrounding, and negative when it is done by the surrounding on the system.

Using this convention, we can see that:
d) when there is a volume compression, the work is done by the surrounding on the system, so the work is negative, and the variation of internal energy is positive due to the negative sign -W in the first law of thermodynamics
e) when heat is added to the system, Q is positive, so the variation of internal energy is positive as well</span>
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Since excited state is 3 fold degenerate

Ne/Ng =3 x e^{(-ΔE)/kt}

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T = 960 K

Constant k = 8.617 x 10^-5 eV/K

Ne/Ng = 3 x e^{-0.25/(8.617x10^-5) x 960}

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           = 3 x 0.0412 = 0.1237 ≅ 0.12

Ne = 0.12 Ng

but Ne + Ng = N, where is N is total number of particles, substituting Ne into equation we get,

Ng(1 + 0.12) = N

Ng = N/1.12 = 0.893N

and Ne = 0.12 x 0.893 N = 0.107 N

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