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

Assignment #3

Carter

Appendix A: 3a,b
Chapter 2: 10
Chapter 3: 1, 2, 3, 5
Chapter 4: 1

Video HW to be turned in separately--see Moodle for deadline

Carter, Chapter 3, problem 8

Notes / Answers

Carter Appendix A
Problem 3

In part b.) find $z(x,y)$ by integrating $dz$ along a specific path from the the origin $(x',y')=(0,0)$ to a final point $(x',y')=(x,y)$ "in $x'$, $y'$ space". In particular, follow the path that runs from $(0,0)\to (0,y)$, and then $(0,y)\to (x,y)$.

Carter Chapter 2
Problem 2-10

One approach is try out this trial equation of state and see if it gives the right $\beta$ and $\kappa$. Consider equations (2.13) and (2.14).

OR: integrate 2.17

Carter Chapter 3
Problem 3-1

For an isothermal process, $T$ is constant. For an ideal gas, $P=\frac 1V nRT$ so, $$\begineq \Delta W &= \int_{V_i}^{V_f}P(V)\,dV\\ &=\int\frac{nRT}{V}\,dV\\ &=nRT\ln \left.V\right|_{V_i}^{V_f}\\ &=nRT\ln\left( \frac{V_f}{V_i}\right). \endeq $$ We are given the initial and final pressures (not the initial and final volumes). Solving the ideal gas equation for $V$: $V=\frac{nrT}{P}$. So, $\frac{V_f}{V_i}=\frac{P_i}{P_f}=\frac{1}{10}$, and the work done is: $$\begineq\Delta W &= nRT\ln(0.1)\\ &=10\text{ kMole }*8.314\times 10^3\text{ J/kmole}*300\text{ K }*(-2.30) &=-5.74\times 10^7\text{ J}\endeq$$

Problem 3-2

Work done is $\int P\,dV$. Since pressure is constant, $$\Delta W = P \Delta V=(30\text{ atm}\cdot 1.01\times 10^5\text{ Pa/atm}) (0.5(0.4/2)^2\pi)\text{[m^3]}=191\text{ kJ}.$$

Problem 3-3

(1 pt) The work done *by* the system is $$\Delta W = \int P\,dV=-\frac{3}{8}nRT_i.$$ So, the work done *on* the system (by the surroundings) is $+\frac{3}{8}nRT_i$.

 

Problem 3-5

You might use CoCalc to do the integral and substitutions in problem 5. Here's a page showing How to substitute values into an expression in Sagemath. And here's one about assumptions: You may want to tell Sagemath to assume certain parameters are, for example, positive, before integrating. Here's a tip about using Assumptions to simplify definite integrals.

Since this is an isothermal process, $T$ is constant when integrating to find the work...

Carter Chapter 3
Problem 3-8

A gas in a frictionless piston. For the path $acb$ (below) 80 J of heat flow into the system, and the system does 30 J of work.


It's useful to write down the internal energy, $U$, at each vertex of the diagram for this problem. You may as well write down $U=0$ at point $a$: This is possible because adding a constant to the internal energy, $U+U_0$, will not change the first law, which only involves changes to the internal energy.

The internal energy, $U$ is a state variable, that depends only on (any) two thermodynamic variables--in this case we can use $P$ and $V$--and not on the path taken.

According to the First Law, a change in the internal energy... $$\Delta U=\Delta Q - \Delta W$$ depends on the heat, $\Delta Q$, that flows in to the system, and the work, $\Delta W$, that the system does on its surroundings.

The initial information that we're given tells us that the difference in internal energy between $a$ and $b$ is: $$\Delta U_{ab}=\Delta Q - \Delta U=80 J - 30 J = 50 J.$$

Carter Chapter 4
Problem 4-1

For part c, calculate the mean heat capacity over that temperature range as (total heat)/(total temperature change). (You'll get a different answer from Carter--who apparently just averaged $c_V$ at the high and low temperatures.)

Correct units for answer are J/kmole/${}^o$K.

Assignment #4

Carter
Chapter 4: 2, 7, 8, 9a, 10, 12, 13

Notes / Answers

Carter Chapter 4
Problem 4-2

b.) One way involves using $c_P=\left(\frac{\del h}{\del T}\right)_P$ and calculating $h$ using (4.21).

Carter Chapter 4
Problem 4-7

Carter, Chapter 4 - problems 8,9 are interrelated. 9a should be straightforward once you have 8a.
Problem 8b

Carter, Chapter 4
Problem 4-10

a.) Show that for this system, an isothermal process is also isobaric. Show that for an isobaric process, $\Delta H=\Delta Q$. Now calculate the enthalpy, $H$.

b.) We're trying to get the relationship between $T$ and $V$. Gee, if we had some expression relating $dV$ to $dT$, which possibly involves functions of $V$, and $T$ as well, we could perhaps use the separation of variables technique to group similar terms and integrate.

Inspired by this thought, consider that we can think of a state function, like $U$, as a function of any two thermodynamic parameters we like. So, think of $U(T,V)$ as a function of $T$ and $V$. Then, the differential of $U$ (the Pfaffian) is: $$dU=\left(\frac{\del U}{\del T}\right)_V dT +\left(\frac{\del U}{\del V}\right)_T dV$$ [Do you have any way of calculating those partial derivatives?]

And, we are supposed to consider an adiabatic process, such that $\delta Q=0$. That might remind you of the first law: $$dU=\delta Q -\delta W=\delta Q- PdV$$ [How can you get $P(T,V)$? Put in your assumption that $\delta Q$=0...]

Now you have two different expressions for $dU$, so you can set them equal to each other...

Luke had a more straightforward approach: Using the expressions given for $U$ and $P$, you can directly calculate the enthalpy: $$H=U+PV=aT^4V+\left(\frac{aT^4}{3}\right)V=\frac 43 aT^4V.$$ Now we consider an initial state with $V_i=V$ and $V_f=2V$, and $T$ is constant. So, we can easily calculate $$\Delta H=H_f-H_i=2*\frac 43 aT^4V-1*\frac 43 aT^4V=\frac 43 aT^4V.$$ That looks like our answer! But wait... It's $\Delta H$, not $\Delta Q$ :-<. What's the connection with heat flow?

From Table 4.2, we have: $$dh=\delta q+v\,dP$$. So $dh=\delta q$ only if the process is isobaric, and we have an isothermal process.

But look at the pressure given to us: $P=aT^4/3$. This means if the temperature is constant, there will be no change in the pressure. So for this system, an isothermal process is also isobaric! And for an isobaric process $\Delta H=\Delta Q$.

Carter, Chapter 4
Problem 4-12 --
b)There are two *different* approaches you can take. In outline form:

1.) [Deeksha's approach] Start from the definition of $H=U+PV$.

1. The differential of enthalpy is $dH=dU+PdV+VdP$.
2. Using the fact that the process is adiabatic, together with the first law, make the argument that for an adiabatic process this simplifies to $dH=VdP$
3. Using our expression $PV^{\gamma}=K$, you can solve for $V(P)$, and integrate $dH$ from initial to final pressure (See 4.12a) to find the change in enthalpy.
4. Some simplification will be required to get the book's answer.

2.) A second approach involves integrating $C_P$:

1. We said that, for an ideal gas, both internal energy, $U$, and enthalpy, $H$, are functions of $T$ alone (no dependence on $P$) and therefore $c_P=\frac{dh}{dT}$ or $C_P=\frac{dH}{dT}$ (not a partial derivative, but just a derivative with respective to one variable).
2. So, you can find the change in enthalpy by integrating $C_P$ (It's a constant--found from Mayer's law...) from initial to final temperature.
3. You can use the ideal gas law and the results of part 4.12a to find the initial and final temperature.

 

Carter, Chapter 4
Problem 4-13

A striking result is that the latent energy of vaporization (538 kCal/kg) is more than half of the total enthalpy change.

You'll need to look up the latent heats for each phase transition.

-1 pt if the latent heats mysteriously appear in your problem writeup without being labelled or explained.

Carter, Chapter 4
Problem 4-14

Use these writing conventions to keep your thinking clear...

• specific (per kilomole) quantitites in lower case,
• total (not per kilomole) quantities in upper case.

I suggest that you do something similar to keep your thinking clear on the difference. For example:

• $\Delta U=U_f-U_i$ is the total change in internal energy,
• We're given the specific heat capacity, $c_v=5R/2$, so $C_v=nc_v$

$P_i= 270$ kPa; $P_f=300$ kPa; $V=V_i=V_f=0.057$ m${}^3$.

Since it's an ideal gas, $U=U(T)$ and $$\Delta U=\int dU=\int_{T_i}^{T_f}C_v\,dT=nc_v(T_f-T_i)=\frac 52nR(T_f-T_i).$$ Using the ideal gas law, $nRT=PV$, so... $$\begineq\Delta U&=\frac 52V(P_f-P_i)=\frac 52(0.057\text{m}^3)(300-270\text{kPa})\\ &= 4.28 kJ \endeq $$