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# Tutor profile: Brian S.

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Brian S.
Experienced Tutor & Freelance Science Writer
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## Questions

### Subject:Writing

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Question:

What kind of writing do I specialize in?

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Brian S.

I have experience in and enjoy both technical and creative writing! I can help you learn to effectively communicate your data in publishable academic papers or distill your ideas into a powerful short fiction story.

### Subject:Calculus

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Question:

Solve $$F = \int_{-\pi\4}^{\pi\4} \frac{1}{2}(3 cos 2\theta)^2 d\theta$$.

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Brian S.

$$F = \frac{9}{2} \int_{-\pi\4}^{\pi\4} (cos 2\theta)^2 d\theta$$ $$2cos^2\frac{\theta}{2} = 1 + cos{\theta}$$ $$F = \frac{9}{4} \int_{-\pi\4}^{\pi\4} (1 + cos4\theta) d\theta$$ $$F = \frac{9}{4}[ \theta + \frac{1}{4}sin4\theta] |_{-\pi\4}^{\pi\4}$$ $$F = \frac{9}{4}[\frac{\pi}{4}+sin(\pi) -(\frac{-\pi}{4}+ sin(-\pi))]$$ $$F = \frac{9\pi}{8}$$

### Subject:Materials Science

TutorMe
Question:

For a free electron gas, find the pressure in three dimensions. Why physically is the pressure of an electron gas at zero temperature not zero, unlike in a classical gas?

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Brian S.

For a Fermi gas of free electrons, the pressure is $$P = -\frac{dU}{dV}|_{S,N}$$ where the total entropy and number of electrons are constant. $$u = \frac{3}{5}nE_F$$; $$E_F = \frac{\hbar^2}{2m}(3\pi^2n)^{2/3}$$ $$U = Vu = \frac{3}{5}NE_F = \frac{3}{5}N \frac{\hbar^2}{2m}(3\pi^2n)^{2/3}$$ $$P = -\frac{dU}{dV}|_{S,N} = \frac{2U}{3V} = \frac{2U}{3}$$ For a classical gas, the pressure is $$P = nkT$$. As T -> 0, P -> 0. For an electron gas, however, we've solved that $$P = \frac{2U}{3}$$. A single quantum state must be filled by only one electron due to the Pauli exclusion principle. As a result, electrons have kinetic energies from zero up to $$E_F$$ at T = 0. This lines up with our math as P > 0 if U >0.

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