# Tutor profile: Gabriela C.

## Questions

### Subject: Physics (Thermodynamics)

Obtain the Maxwell relations from the differential form of the thermodynamic potentials.

The thermodynamic potentials are: $$dU=TdS-pdV$$ $$dH=TdS+Vdp$$ $$dF=-SdT-pdV$$ $$dG=-SdT+Vdp$$ These expressions are exact differentials. When we have a function $$f(x,y)$$ whose exact differential is: $$df=M(x,y)dx+N(x,y)dy$$ then it follows that: $$\big(\frac{\partial{M}}{\partial{y}}\big)_x=\big(\frac{\partial{N}}{\partial{x}}\big)_y$$ Knowing this, we can apply this property in each of the thermodynamic potentials, obtaining: $$\big(\frac{\partial{T}}{\partial{V}}\big)_S=-\big(\frac{\partial{p}}{\partial{S}}\big)_V$$ $$\big(\frac{\partial{T}}{\partial{p}}\big)_S=\big(\frac{\partial{V}}{\partial{S}}\big)_p$$ $$\big(\frac{\partial{S}}{\partial{V}}\big)_T=\big(\frac{\partial{p}}{\partial{T}}\big)_V$$ $$-\big(\frac{\partial{S}}{\partial{p}}\big)_T=\big(\frac{\partial{V}}{\partial{T}}\big)_p$$ These equations are known as the Maxwell relations.

### Subject: Physics

Calculate the escape velocity from Earth for a rocket (suppose friction effects are null). $$G = 6.67 \times 10^{-11} \frac{Nm^2}{kg^{-2}} , R_E = 6.37 \times 10^6 m , M_E = 5.98\times 10^{24} kg$$

By conservation of energy, we know it has to be constant at every position of the spacecraft. Let's suppose then, the spacecraft "reaches infinity", so the Earth will no longer pull from it (it would stop), the mechanical energy then becomes zero. Our equation takes the form: $$E_m=E_k+E_p=0$$ Where $$E_k, E_p$$ are the kinetical and potential energy respectibly. Explicitly, we can write the equation above as: $$E_k+E_p=0$$ $$\frac{1}{2}m_sv_e^2-\frac{GM_Em_s}{R_E}=0$$ Knowing the Earth's mass and radius, the gravitational constant and the spacecraft mass, we can obtain the escape velocity from the spacecraft: $$v_e^2=\frac{2GM_Em_s}{R_Em_s}$$ $$v_e=\sqrt{\frac{2GM_E}{R_E}}$$ $$v_e=\sqrt{\frac{2(6.67 \times 10^{-11} \frac{Nm^2}{kg^{-2}})(5.98\times 10^{24} kg)}{6.37 \times 10^6 m}}$$ $$v_e=11191\frac{m}{s}\approx 11.2\frac{km}{s}$$ Note that this velocity does not depend on the object's mass.

### Subject: Algebra

How can I solve the equation listed below? $$x^2+x-6=0$$

Using the cuadratic formula: $$x=\frac{-b^2 \pm \sqrt{b^2-4ac}}{2a} $$ we identify the $$a, b, c$$ terms, each one corresponding the power of x. In this case: $$a=1$$ $$b=1$$ $$c=-6$$ Substituying this values in the equation shown above: $$x=\frac{-(1)^2 \pm \sqrt{(1)^2-4(1)(-6)}}{2(1)}$$ $$x=\frac{-1 \pm \sqrt{1+24}}{2}$$ $$x=\frac{-1 \pm 5}{2}$$ We get the two values of $$x$$ by evaluating both signs of the square root: $$x_1=\frac{-1 + 5}{2}=\frac{4}{2}=2$$ $$x_2=\frac{-1 - 5}{2}=\frac{-6}{2}=-3$$

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