Download Physics 4230 Set 2 Solutions Fall 1998 Fermi 2.1) Basic 1st Law of

Physics 4230
Set 2 Solutions
Fall 1998
Fermi 2.1) Basic 1st Law of Thermodynamics:
Calculate the energy variation of a system which performs 3.4x108 ergs of work and
absorbs 32 calories of heat.
So, the bottom line in this problem is whether you can remember the 1st Law and
whether you get the signs right. 1st things first. The Law says that the internal energy of a
system can change either because work got done or because heat got transferred. The
standard form we use is that:
where a positive W means work is done on the system and positive Q means heat went
into the system. Fermi’s problem says that his system performs work. In other words, it
did work on something else. Therefore, the work done on the system is -3.4x108 ergs. The
heat absorbed is fine regarding sign, but needs to be converted to units consistent with the
work (or you could do it the other way, or convert both to some other energy unit like
Joules). Using Fermi Eq. (19), 1 calorie = 4.19x107 ergs, we have a heat input to the
system of 1.34x109 ergs. Thus,
Therefore, for this system, we pumped in more heat than the system lost in doing the
work and the internal energy goes up.
Fermi 2.2)
1st Law used to find the heat:
How many calories are absorbed by 3 moles of an ideal gas expanding isothermally from
the initial pressure of 5 atm. to the final pressure of 3 atm., at the temperature of 0 C?
Again, we must get the work, the internal energy change, and the signs correct. Once you
do that, then determining the heat (aside from units conversion to calories) is easy:
Just to be certain that we recognize the work piece, the system is setup by Fermi so that
the change is isothermal at 0 C (273.15 K) and involves expansion of the gas.
Physics 4230
Set 2 Solutions
Fall 1998
Therefore, the gas is doing work on the outside and is going to be a negative. The PV
diagram for the process is just the standard isothermal line:
The work is given by the area under the isotherm (with sign adjusted to agree with the
direction of travel on the path). Of course, Fermi already did this integral for Eq. (10) or you
could check out the solutions to Problem Set 1. In any case, the work done is just:
which yields a final work done by the system of W=3.4x 1010 ergs. The energy change in the
system is zero for this case because the system is an ideal gas and therefore has an internal
energy change given by:
which, since we are on an isotherm, is simply an energy change of zero. Therefore, the work
done and the heat absorbed are equal:
Physics 4230
Fermi 2.3)
Set 2 Solutions
Fall 1998
1st Law Work and heat:
One mole of a diatomic ideal gas performs a transformation from an initial state for
which temperature and volume are, respectively, 291 K and 21,000 cc. to a final state in
which temperature and volume are 305 K and 12,700 cc. The transformation is
represented on the (V,P) diagram by a straight line. Find the work performed and the heat
absorbed by the system.
Once again, we have a situation where the system evolves along some path in the PV
plane and we are asked for the work and the heat associated with the path. Once again,
let’s start by drawing a picture of the path in the PV plane. We are given the initial and
final volumes, but we still need the associated pressures to draw the line. Plug and chug
the conditions through the ideal gas equation of state to get:
Then, the PV diagram looks like:
To find the work, we must integrate the area under the line and to do that, we must know
the equation of the line. Therefore, we need the slope, a, and the intercept, b:
Both are easily found and are:
Physics 4230
Set 2 Solutions
Fall 1998
The integrated area is then the work. Note that Fermi asks for the work performed by the
system. However, it is clear that the gas is being compressed, so that work is actually being
done on it Therefore, we expect that the work performed by the gas will be a negative
number:
Plugging in the values for a and b then yields, for the work performed:
Next, Fermi asks for the heat absorbed. The short route to find the heat is again by knowing
the change in internal energy of the gas and then subtracting out the work done on the gas.
Note that the work done on the gas is +1.3kJ:
SO, the name of the game is to find the change in internal energy. One way to do this is to
parallel my discussion in class and re-derive the internal energy by calculating it along
special paths where you can use and CV. I will simply note that Fermi has done this for you in
Fermi Eq. (29). It says that:
For the diatomic ideal gas, CV = 5NkB/2 so that:
In generating this result, I have used the fact that we have one mole of gas so that the number
of particles is Avogadro’s number, NA, and that Boltzmann’s constant and the ideal gas
constant are just related by Avogadro’s number. The heat is then simply:
Physics 4230
Set 2 Solutions
Fall 1998
Fermi 2.4) Adiabatic expansion:
A diatomic gas expands adiabatically to a volume 1.35 times larger than the initial
volume. The initial temperature is 18 C. Find the final temperature.
This is a cut and dried use of Fermi Eq. (38):
for adiabatic expansion, with some experimental results for the ratio of specific heats:
Thus, we simply have that:
or, by rearranging sides and plugging in the ratio of specific heats and the prescribed
volume ratio, we have:
Physics 4230
Set 2 Solutions
Fall 1998
Plotting of PV lines.
Here, I just wanted to see decent plots of the PV lines which you had to use in Fermi 2.3
and 2.4. I didn’t necessarily want you to draw isotherms unless you actually used them in
the solutions.
My major point here is to try to make you internalize the connection between the picture
of the PV plane and the mathematics used to find the work. Many people, for example
had no difficulty drawing a straight line on the PV diagram for Fermi 2.3, but when they
tried to calculate the work, couldn’t figure out what function to use for P(V)! This,
despite the fact that they could draw it!. SO the idea is that once you are forced to plot the
function enough times (which requires that you actually write the function into
Mathematica) eventually the connection between the picture and the math will be forged.
Nobody had any real trouble plotting, just in picking the functions to plot.
Problem 2) Equation of State for Solids
The approximate equation of state I asked you to describe is:
V = V0 + aP + bT
Here’s what I was looking for:
1) V0 is the volume the solid has at zero pressure and temperature.
2) The quantity a is the change in volume with pressure. It ought to be negative.
3) The quantity b is the change in volume with temperature. It could be either sign.