# Tutor profile: Georgios A.

## Questions

### Subject: Physics (Thermodynamics)

An ideal gas undergoes an isothermal process, while it expands to twice its volume. Calculate the change in the entropy for this process. Is your result consistent with what you expected to find?

The change in entropy is given by: $$\Delta S=\int\frac{dQ}{T}$$, where $$Q$$ stands for the heat exchanged. For an isothermal process the change in internal energy is zero, so from the 1st law of thermodynamics: $$dU=dQ-PdV$$, so $$dQ=PdV$$. Then, $$\Delta S = \int \frac{P}{T}dV$$. Both pressure and temperature depend on volume through the eq. of state $$PV=nRT\Rightarrow \frac{P}{T}=\frac{nR}{V}$$, so $$\Delta S=nR\int\frac{dV}{V}=nR\ln\left(\frac{V_2}{V_1}\right)$$, or $$\boxed{\Delta S =nR\ln2>0}$$ As expected, when the gas expands the entropy increases.

### Subject: Physics (Newtonian Mechanics)

A point object of mass $$m$$ is placed at rest, in an area where the force is time-dependent and given by $$\vec{F}(t)=(2at+b)\hat{x}$$, where $$a,\,b$$ are non-zero constants. Derive the expression that describes the object's velocity.

In order to determine the object's path, we need to write down Newton's eq. and since the force points to only one dimension (x), it is sufficient to write the eq. in its one-dimensional format: $$m\ddot{x}=F_x$$, where $$F_x=(2at+b)$$ Now we can integrate Newton's eq. and find: $$m\dot{x}=\int dt(2at+b)=at^2+bt+c$$, where $$c$$ is a constant of integration. To determine it, remember that $$v(t=0)=0$$. The velocity is $$v(t)=\dot{x}=m^{-1}(at^2+bt+c)$$, so $$v(0)=c=0$$. Therefore, $$\boxed{v(t)=(at^2+bt)/m}$$

### Subject: Physics (Electricity and Magnetism)

Consider the first Maxwell eq. in differential format $$\vec{\nabla} \cdot \vec{E}=\rho/\epsilon_0$$. What is the equivalent integral format? Express the final result with respect to the electric flux?

First of all, let us make clear what is meant by integral form. The equation provided above contains the differential operator $$\vec{\nabla}$$ which acts on the electric field. Thus, the characterization "differential format". It is only reasonable that in order to derive the integral one, we need to implement integrals in our analysis. We can merely, act on both sides of the eq. with the integral operator $$\int d^3x$$, assuming that the field configuration is in the 3-d space. Then: $$\int d^3x \,\vec{\nabla} \cdot \vec{E} = \int d^3x\, \rho/\epsilon_0.$$ The second-hand side is nothing more but the electric charge contained in the region we integrate with respect to. To see that remember that $$d^3x\equiv dV$$, where $$V$$ is the volume and also that $$\rho=dq/dV$$, where $$q$$ is the electric charge. Now, $$\int dV \,\vec{\nabla} \cdot \vec{E} =q/\epsilon_0.$$ Even though this is an integral eq. we are not done yet. Note that we are asked to express our result in terms of the electric flux, namely $$d\Phi = \vec{E}\cdot dA$$, where $$\vec{A}$$ is the vector that corresponds to the surface the electric field goes through. In order to make the flux appear, we need to make the volume integral on the left-hand side, a surface integral. At this point, we make use of the divergence theorem according to which: $$\int dV \,\vec{\nabla} \cdot \vec{E} =\oint d\vec{A}\cdot \vec{E}=\Delta\Phi.$$ The surface is the one that encloses the volume with respect to which we integrated in the first place. Therefore finally, $$\boxed{\Delta\Phi=q/\epsilon_0}$$

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