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physics question help please

hi, please could i have help on part d of this question? I used the equation U=3/2NKbT and subbed in the temperature of 18 and -10 (after converting to kelvin.) Then I subtracted both U i calculated to find the change in internal energy but I'm getting nowhere near the answer in the ms?
Thanks!Screenshot 2025-09-09 221435.png
ms.png

Reply 1

Original post
by anonymous56754
hi, please could i have help on part d of this question? I used the equation U=3/2NKbT and subbed in the temperature of 18 and -10 (after converting to kelvin.) Then I subtracted both U i calculated to find the change in internal energy but I'm getting nowhere near the answer in the ms?
Thanks!Screenshot 2025-09-09 221435.pngms.png

The way I’d approach that question is by equipartition theorem (probably a bit overkill, but it looks like it should work).

Given helium is a monatomic gas, it has three translational degrees of freedom (and no rotational or vibrational degrees of freedom to consider, since it’s monatomic - thank goodness!).

Equipartion theorem states that each of these degrees of freedom contributes 1/2 kT to the kinetic energy of each helium particle. That is to say at the temperature T, the total kinetic energy of each helium atom will be 3/2 kT. I’ll even go further and assume there is no potential energy to worry about given we are assuming it to be an ideal gas (e.g. no interactions between particles) and thus the kinetic energy must be the total internal energy at the given temperature.

Now plugging in the numbers:

At -10°C (263.15 K), U = 5.45… x 10^-21 J
At +18°C (291.15 K), U = 6.03… x 10^-21 J

Difference 5.8 x 10^-22 J

So presumably the question meant to ask for the change in internal energy for a single particle of helium and not the change in internal energy for the entire sample.
(edited 2 months ago)

Reply 2

Original post
by anonymous56754
hi, please could i have help on part d of this question? I used the equation U=3/2NKbT and subbed in the temperature of 18 and -10 (after converting to kelvin.) Then I subtracted both U i calculated to find the change in internal energy but I'm getting nowhere near the answer in the ms?
Thanks!Screenshot 2025-09-09 221435.png
ms.png


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