# 9.7: Problems

• • Contributed by Marcia Levitus

Problem $$\PageIndex{1}$$

Determine whether the following differentials are exact or inexact. If they are exact, determine $$u=u(x,y)$$.

1. $$du=(2ax+by)dx+(bx+2cy)dy$$
2. $$du=(x^2-y^2)dx+(2xy)dy$$

Problem $$\PageIndex{2}$$

Determine whether dz is exact or inexact. If it is exact, determine $$z=z(P,T)$$.

$dz=-\frac{RT}{P^2}dP+\frac{R}{P}dT \nonumber$

Problem $$\PageIndex{3}$$

From Equation \ref{eq:dG}, and using the fact that $$G$$ is a state function, prove that the change in entropy ($$\Delta S$$) of one mole of an ideal gas whose pressure changes from an initial value $$P_1$$ to a final value $$P_2$$ at constant temperature is:

$\Delta S =-R \ln{\frac{P_2}{P_1}}\nonumber$

Problem $$\PageIndex{4}$$

From Equations \ref{eq:dU}-\ref{eq:dA}, and using the fact that $$U,H$$ and $$A$$ are state functions, derive the three corresponding Maxwell relations.

Add texts here. Do not delete this text first.

Problem $$\PageIndex{5}$$

Given the following differential:

$dz=xy dx + 2y dy\nonumber$

1. Determine if it is exact or inexact. If it is, obtain $$z(x,y)$$
2. Calculate the line integrals $$\int_c{dz}$$ for the paths enumerated below:
1. the line $$y=2x$$ from $$x=0$$ to $$x=2$$
2. the curve $$y = x^2$$ from $$x = 0$$ to $$x = 2$$
3. any other path of your choice that joins the same initial and final points.

Problem $$\PageIndex{6}$$

For a mole of a perfect monoatomic gas, the internal energy can be expressed as a function of the pressure and volume as

$U = \frac{3}{2}PV\nonumber$

1. Write the total differential of $$U$$, $$dU$$.
2. Calculate the line integrals $$\int_c{dU}$$ for the paths shown below ($$c_1, c_2, c_3$$): 3. Calculate $$U(V_f,P_f)-U(V_i,P_i)$$ and compare with the results of b) (Note: $$f$$ refers to the final state and $$i$$ to the initial state).
4. Considering your previous results, calculate $$\int_c{dU}$$ for the path below: As defined in Section 9.3,

$\label{eq:dU} dU=T(S,V)dS-P(S,V)dV$

$\label{eq:dP} dH=T(S,P)dS+V(S,P)dP$

$\label{eq:dA} dA=-S(T,V)dT-P(T,V)dV$

$\label{eq:dG} dG=-S(T,P)dT+V(T,P)dP$