Since Carter has many answers in the back of the book, you must show your work to receive credit for your answers.

Assignments: 1 | 2

Assignment #1

Carter, Chapter 1
Chapter 1: 1-6 (that is, problems 1 through 6), and 8

Helrich
2.7: Let's assume an idealized atmosphere: The atmosphere is all at the same temperature, $T$, and the air obeys the ideal gas law, $PV=nRT$, exactly.

Consider a column of air with a constant horizontal cross-sectional area, $A$. The variable $h$ indicates the height in the column.

A small chunk of air has a volume $dV = A\,dh$. In terms of its density $\rho$, the mass of this chunk of air is $dm = \rho A \, dh$. Let $M$ be the atomic weight (kg / kmol) of the gas. The weight of this mass of air is supported by the difference in pressures $dP=P_2-P_1\equiv P(h)-P(h+dh)$ between the bottom of $dV$ (higher pressure, $P_2$) and the top. (The height of the chunk is $dh$). Come up with a differential expression for $dP$ in terms of $dh$, and then integrate it from $h_0$ to $h$ to show that the variation of pressure with altitude for such an idealized atmosphere is: $$ P(h)=P_0 \exp\left(\frac{-Mg(h-h_0)}{RT}\right).$$

Notes / hints

Helrich, Problem 2.7: Using $\color{red}{{}^*}$: $$PV=nRT.$$

Hey wait a minute, maybe we should be switching m and M to match our conventions for extensive and intensive variables. Or is this right?

  1. Consider a very short (height, $dh$) chunk of air.
  2. Draw a force diagram with all the forces acting on that chunk of air, so as to keep the chunk of air stationary. This will give you a relationship between $dh$ and $dP$--the pressure difference between the air above the chunk and below the chunk. It also likely involves the density of the gas, which is *also* a function of $h$! At the point of getting a relationship between $dh$ and $dP$, check your signs: Think of $dP$ as a positive number and ask yourself if pressure $P$ must increase or decrease with positive changes, $dh$, in height in order to make the forces on our small chunk of air sum to zero?
  3. Use the ideal gas law to find the density of your gas, at whatever the constant temperature is. With
    • $n$=# of kilomoles of gas,
    • $m$=# of kg of gas, and
    • $M$=# of kg/kilomole of gas (molecular weight of the gas),

    we can write the gas law as: $$PV=nRT=\frac{m}{M}RT.$$

  4. Now use the technique of separation of variables to gather $dP$ and all functions of $P$ on one side of the equation, and $dh$ and any functions of $h$ and constants on the other side. Now you can integrate both sides.
  5. After integrating, use what you know about logarithms and exponentials to solve the resulting expression for $P(h)$.

$\color{red}{{}^*}$ - Aack! Sorry folks! Carter writes things this way in section 2.2 of his textbook--probably because folks have been using $n$ for so long to stand for the amount of substance (in a system). But it's at odds with the standards for how you name thermodynamic variables as extensive or intensive.

Carter, Problem 1-4:

Me: 170 lbs. Area of 2 feet 10 X 30 X 2 =600 cm^2=0.06 m^2
$P=F/A=(77$kg$*9.8 m/s^2)/0.06$m${}^2=$0.13 atmospheres = 13,000 Pa = 96 Torr.

Area of heels is, say, 1/5 of shoes. Multiply each pressure above by 5: ~0.65 Atm = 65,000 Pa = 500 Torr.

Problem 1-5: You can use CoCalc to solve systems of equations. For example, if you know 2 equations: $$3a+b=14; \ \ \ \ a-b=2$$ for two unknown quantities $a$ and $b$, you can solve them using this syntax:

a,b=var('a,b')
solve(
 ( 3*a + b == 14 ,  a - b == 2 ),
  (a, b)
 )

We're told that resistance as a function of temperature is $$R=aT+b$$, and that in a certain scale $R=5$ ohms, $T=100{}^o$, and that when $R=6$, $T=500{}^o$.

Here's the quick CoCalc way of solving the two resulting equations:

Solving the two equations for $a$ and $b$, the resistance in ohms is given by: $$R(T)=\frac 1{400}T+4.75$$. Solving for $T$ with $R=$5.4 ohms, $T=260{}^o$ on this scale.

Problem 1-6: You need to use equation (1.18) to find $T$ from the $P_{TP}/P$ ratio. You can graph the $P_{TP}/P$ ratios vs. $P_{TP}$, fit the data to a line, and use the $y$-intercept as your estimate of the limit as $P_{TP}\to 0$. Probably the quickest way to fit data to a line is to use Desmos. See this example (an intro to Desmos).

Conceptually, it is simpler to use equation 1.18 to calculate an estimate of $T$ at each $P_{TP}$ value. Then you can graph to find $T$ vs $P_{TP}$ and fit the resulting line to find the limit.

Include a legible URL to your fit in Desmos. And adjust the $y$ limits to show the fit as a sloping line.

A somewhat more complicated approach is to do the fitting in Sagemath. Here is a Jupyter notebook, Lab01.ipynb which you can upload to CoCalc and modify. Here is a pdf version Lab01.pdf.



Or, using CoCalc

Jaime had a cool approach (below): He calculated the temperature for each data point, and took the limit of that as $P_{TP}\to 0$. The temperature approximation is exactly proportional to the ratio of pressures, and it has Desmos doing the calculation of the temperature directly.
Also shown are two other attempts to calculate the y-intercept.

  • The second approach takes off from the observation that the function is more linear the closer $P_{TP}$ is to zero, so I only used the closest two points to estimate the intercept.
  • The third approach is to fit to a quadratic function.
But none of these matches Carter's answer of 348.65 K.

Problem 1-8

Solve two equations for $a$ and $b$. If you find more than one solution, justify which one you picked.

Assignment #2

Review line integrals. For example, from Paul's MVCalculus course, section 13.2.

Carter
Appendix A: 2a,b (See example of contour integration from Phys 303) See also example problem A1 below
Chapter 2: 2, 3, 7, 9

Notes on HW 2 / Solutions

A1

part a: $dz= 2x\ln y\,dx + (x^2/y)\,dy \equiv M(x,y)\,dx+N(x,y)\,dy$.

The test for exactness is $\frac{\del M}{\del y} \stackrel{?}{=} \frac{\del N}{\del x}$. In this case we find both partial derivatives are $2x/y$, so $dz$ is indeed exact. See the screencast for a bit more:

A1-a - test for exactness (2.5 min).

To find $z(x,y)$, you can either:

  1. guess a particular $z$, and then differentiate to show that you get the right partial derivatives, or else,
  2. Integrate $dz$ from some arbitrary reference point (often the origin) to $(x,y)$.

For this $dz$, since it involves $\ln(y)$, and $\ln(0)$ is undefined, you should definitely *not* pick the origin! Instead, pick a different starting point than the origin. (Say (0,1), or more generally $(x_0,y_0)$).

Integrating $dz$ to find $z(x,y)$ (7 min). $$z(x,y)=x^2\ln y+C.$$

Carter Appendix A
Problem 2

For full credit: Show how you integrate to find $z(x,y)$. For partial credit: Guess $z(x,y)$ and check that your guess is correct.






Carter, Problem 2-2

In order to get an answer in kilomoles, make sure you're using $R=8.314\times 10^3 $ J/kmol/K. Also temperature must be in degrees Kelvin (= deg C + 273).

1 Pascal = 1 N/$m^2$=1 N$\cdot m$/$m^3$=1 J/$m^3$.

  1. Solve ideal gas law for $n$: $$n=\frac{P V}{R T}=\frac{(1.5\times 10^6\text{ Pa})\ 0.5\text{ m}^3} {(8.314\times10^3\text{J / kmole}\cdot K)(20+273)K}=0.308\text{ kmole}$$
  2. Oxygen gas is $O_2$, so its molecular weight is ~ 32 kg/kmole. Multiplying molecular weight by kilomoles gives 9.85 kg.
  3. Since $n$, $V$, and $R$ are constant, $P/P_0=T/T_0=(500+273)/(20+273)=2.64$. So $P=2.64*P_0=4.0\times 10^6$ Pa.
  4. Answer is 90% of part a=0.308*.9=0.28 kilomoles. (This is the amount that can be removed).


Carter, Chapter 2, Problem 3

Hint: You can find $A$ in terms of the initial $P_1$, $T_1$ (and not $v_1$) by using the ideal gas law.





Carter Chapter 2
Problem 7

You *can* solve three equations in three unknowns in CoCalc. To find the critical pressure, volume and temperature you would solve for $P,v,T$ these three equations:

  1. The first equation is the Dieterici equation itself, $P=....$.
  2. The second equation is that the first derivative of $P$ w.r.t. $T$ vanishes at the critical point, that is $\left(\frac{\del P}{\del v}\right)_T=0$,
  3. And the third equation is that the second derivative of $P$ vanishes.
Perhaps you will code these up and try something like this in SageMath:
    solve( [eq1, eq2, eq3], P,v,T )

And will get a ton of error messages!

Look back at that system of equations: The second and third equation will not actually contain any factors of $P$. So, if you were solving this by hand it would make sense to use the second two equations to solve for $v$ and $T$ first, and finally use the first equation to find $P$.

Perhaps this is why, if you change the order of variables in your solve() command, placing $P$ last:

    solve( [eq1, eq2, eq3], v,T,P )

...Now CoCalc will deliver unto you a tidy solution without the tiniest squeak!

Here's a pdf of solving this in CoCalc
and in Mathematica.

Carter Chapter 2
Problem 9

Take the partial derivatives using the definitions 2.13 and 2.14. You'll have some 2nd order cross differentials.

There are two common blind alleys:

  • You can show the relationship holds for ideal gases by using the ideal gas law to calculate partial derivatives such as $(\del v/\del T)_P=R/P$. But I want you to show (and you *can* show) that the relation holds universally, and not just for ideal gases. You do this by using the definitions of $\beta$ and $\kappa$, (2.13) and (2.14).
  • Using the definition of $\beta$ we can write: $$\begineq \left(\frac{\del \beta}{\del P}\right)_T&=\left(\frac{\del}{\del P}\left[\frac1v\left(\frac{\del v}{\del T}\right)_P\right]\right)_T \endeq $$ It is tempting to pull the factor $1/v$ out of the partial derivative to get $\frac 1v \left(\frac{\del}{\del P}\left(\frac{\del v}{\del T}\right)_P\right)_T$. But that would be wrong because both $1/v$ and $\del v/\del T$ might depend on $P$. So, instead you must treat the contents of $[...]$ as the product of two functions of $P$, and use the product rule when you take the partial derivative wrt $P$.