Thermo Thursday returns. But on a Tuesday! WHO COULD HAVE SUSPECTED?

This section introduces isothermal and adiabatic compression. Isothermal compression is compression that occurs so slowly that no heat is added to the gas. For isothermal compression of an ideal gas, the temperature remains constant, so you can use the ideal gas law, PV=NkT, and the integral equation for work done during compression, W=-\int _{{V}_{i}}^{{V}_{f}}{P(V)dV}, to derive that {W}_{isothermal}=NkT\ln {\frac {{V}_{i}}{{V}_{f}}}. Since work is being done but the temperature is not changing, heat must be flowing out of the gas, in an amount equal to the work being done. On a PV graph, an isothermal compression takes the shape of a concave-up hyperbola.

Adiabatic compression, in contrast, happens so fast that no heat escapes from the gas during the process. Thus, for adiabatic compression, \Delta U = W.  The PV curve for this starts on a lower temperature isotherm and ends on a higher temperature isotherm.  To find an equation for the shape of that curve, we look at the equipartition of energy theorem, U = \frac {f}{2}NkT, where f is the number of degrees of freedom per molecule. The infinitesimal change in energy along the curve is then given by dU=\frac {f}{2}NkdT. If we assume the compression is quasistatic, then from the equation for work we know that dU=-PdV. (We can say this because we previously established that in adiabatic compression the entire change in energy comes from work.) This gives \frac {f}{2}NkdT=-PdV. Now you can plug in the ideal gas law for P and do some canceling to get \frac {f}{2}\frac {dT}{T}=-\frac {dV}{V} \rightarrow \frac {f}{2}\ln {\frac {{T}_{f}}{{T}_{i}}}=-\ln{\frac{{V}_{f}}{{V}_{i}}}.  This ends up simplifying down to V{T}^{{f}/{2}}=C for some constant C. If you are looking for pressure instead of temperature, you can use the ideal gas law to rewrite this {V}^{\gamma}P=D for some constant D, where \gamma = {(f+2)}/{f} and is called the adiabatic constant.

Problem 1.35: Derive {V}^{\gamma}P=constant from V{T}^{{f}/{2}}=constant.

I’m going to use a subscript on the constant term to show when that side of the equations changes. Let’s start with V{T}^{{f}/{2}}={C}_{0}. I want to get this equation in terms of pressure, so we use the ideal gas law to say that T=\frac {PV}{Nk}. This gives us
V{(\frac{PV}{Nk})}^{{f}/{2}}={C}_{0}.
Raising both sides of the equation to the power 2/f  and moving the constant terms to the right side gives
{V}^{{2}/{f}}VP={C}_{1}.
I can then combine the volume terms,
{V}^{1+\frac{2}{f}}P={V}^{{(f+2)}/{f}}P={V}^{\gamma}P={C}_{1}.

Problem 1.36: In the course of pumping up a bicycle tire, a liter of air at atmospheric pressure is compressed adiabatically to a pressure of 7 atm. (Air is mostly diatomic nitrogen and oxygen.)

(a) What is the final volume of this air after compression?

I will treat the air as a diatomic ideal gas for these calculations, which means the air molecules have five degrees of freedom. This gives me \gamma =7/5. For adiabatic compression, {V}_{i}^{\gamma}{P}_{i}={V}_{f}^{\gamma}{P}_{f}, so
{V}_{f}={(\frac{{V}_{i}^{7/5}{P}_{i}}{{P}_{f}})}^{5/7}
which for an initial pressure of 1 atmosphere and volume of 1 liter, and final pressure of 7 atmospheres, gives a final volume of 0.25 liters.

(b) How much work is done in compressing the air?

The total change in energy is the heat added or lost plus the work done, \Delta U = Q + W. Since this is adiabatic compression, there is no heat lost to the environment, so the entire change in energy is due to work. From the equipartition of energy theorem and the ideal gas law I can write \Delta U= \frac {f}{2} \Delta (PV), which for this system gives
\Delta U = \frac {5}{2} ({P}_{f}{V}_{f}-{P}_{i}{V}_{i})
Plugging in {V}_{f}=0.25 L, {P}_{f}=7 atm, {V}_{i}=1.0 L, {P}_{i}=1 atm, converting to SI units, and calculating gives an energy added due to work of 189.5 J.

(c) If the temperature of the air is initially 300 K, what is the temperature after compression?

I can use the equation V{T}^{{f}/{2}}=constant to say that
{V}_{i}{T}_{i}^{\frac{5}{2}}={V}_{f}{T}_{f}^{\frac{5}{2}}
and so
{T}_{f}={(\frac {{V}_{i}{T}_{i}^{\frac{5}{2}}}{{V}_{f}})}^{\frac {2}{5}}.
Plugging in the values and calculating this out gives a final temperature of 522 K.

Problem 1.37: In a Diesel engine, atmospheric air is quickly compressed to about 1/20 of its original volume. Estimate the temperature of the air after compression, and explain why a Diesel engine does not require spark plugs.

Since the air is compressed “quickly,” I will assume adiabatic compression, and since it’s air that’s being compressed it has the same degrees of freedom as the previous problem. So I can use the last equation from part (c) above, plug in {V}_{f}=\frac {1}{20}{V}_{i}, and simplify to get
{T}_{f}={(20{T}_{i}^{\frac {5}{2}})}^{\frac {2}{5}} \approx 3.3{T}_{i}.
So if the initial temperature of the air is 300 K, then after compression the temperature will rise to around 990 K. Since the ignition temperature of Diesel fuel is 483 K, the air after compression is hot enough to ignite it without a spark.

Problem 1.38: Two identical bubbles of gas form at the bottom of a lake, then rise to the surface. Because the pressure is much lower at the surface than at the bottom, both bubbles expand as they rise. However, bubble A rises very quickly so not heat is exchanged between it and the water. Meanwhile, bubble B rises slowly (impeded by a tangle of seaweed), so that it always remains in thermal equilibrium with the water (which has the same temperature everywhere). Which of the two bubbles is larger by the time they reach the surface? Explain your reasoning fully.

Bubble A undergoes adiabatic compression (in this case, expansion), while bubble B undergoes isothermal compression. Initially, the bubbles are identical, so their pressures and volumes are equal. For isothermal compression, {P}_{B}{V}_{B}=C where C is a constant, and for adiabatic compression {P}_{A}{V}_{A}^{\gamma}=C. The constants C must be the same, because the bubbles are initially identical. Since \gamma > 1, {V}_{A}<{V}_{B}. Thus, the bubble that undergoes isothermal compression, bubble B, is larger when it reaches the surface. This can be seen by visual inspection of a graph depicting an isotherm and an adiabat. Notice that for the same initial conditions, the volume of the isotherm rises faster than that of the adiabat.

It’s the end of the semester and I’ve got other projects I need to be working on, so this week I’m just going to do the last problem from the compression work section and leave the adiabats and isotherms for next time.

Problem 1.34: An ideal diatomic gas, in a cylinder with a movable piston, undergoes the rectangular cyclic process shown in the figure to the right. Assume that the temperature is always such that rotational degrees of freedom are active, but vibrational modes are “frozen out.” Also assume that the only type of work done on the gas is quasistatic compression-expansion work. (a) For each of the four steps A through D, compute the work done on the gas, the heat added to the gas, and the change in the energy content of the gas. Express all answers in terms of {P}_{1}, {P}_{2}, {V}_{1}, and {V}_{2}. (Hint: Compute \Delta U before Q, using the ideal gas law and the equipartition theorem.)

Step A: This is a diatomic gas with no vibration modes, so for this system the equipartition theorem says that U = \frac {5}{2}PV. For this step, \Delta U = \frac{5}{2}({P}_{2}-{P}_{1}){V}_{1}. Since the volume of the gas doesn’t change in this step, there is no work being done on the gas. Recalling that \Delta U = W + Q, that means that Q = \frac{5}{2}({P}_{2}-{P}_{1}){V}_{1}.

Step B: here \Delta U = \frac{5}{2}{P}_{2}({V}_{2}-{V}_{1}). The volume is increasing, so the work done on the gas is negative, W=-{P}_{2}({V}_{2}-{V}_{1}). Since W is negative, that means that the heat added must be the total energy change plus the amount of energy subtracted by negative work. Q=\Delta U - W = \frac{7}{2}{P}_{2}({V}_{2}-{V}_{1}).

Step C: this time \Delta U = \frac{5}{2}({P}_{1}-{P}_{2}){V}_{2}. Note that this is a negative quantity, the system has lost energy. Again, there is no volume change, so no work is being done on the gas. Thus the entire energy change is due to the system losing heat energy, so Q=\frac{5}{2}({P}_{1}-{P}_{2}){V}_{2}.

Step D: in the last step, \Delta U = \frac{5}{2}{P}_{1}({V}_{1}-{V}_{2}). This is another negative quantity. But this time, the work being done on the gas is positive, W=-{P}_{1}({V}_{1}-{V}_{2}). Since the work done on the gas is positive, and the energy change is negative, it must be losing even more heat energy than it is gaining in work energy. Here Q= \Delta U - W = \frac{7}{2}{P}_{1}({V}_{1}-{V}_{2}).

(b) Describe in words what is physically being done during each of the four steps; for example, during step A, heat is added to the gas (from an external flame or something) while the piston is held fixed.

The author went ahead and did step A for me. In step B, the volume of the gas increases, so the piston is being drawn out (or, more likely, pushed out by internal pressure). But this doesn’t result in any decrease in pressure, which means that the gas is also being heated as the volume increases. In step C, the system just loses heat energy, so the system is being cooled while the piston is held fixed. In step D the piston is compressed, but the cylinder is still being cooled so the pressure doesn’t change.

(c) Compute the net work done on the gas, the net heat added to the gas, and the net change in energy of the gas during the entire cycle. Are the results as you expected? Explain briefly.

The net change in energy is zero, that’s what makes it a cycle. For the work over the whole cycle, that’s {W}_{cycle}=-({P}_{2}-{P}_{1})({V}_{2}-{V}_{1}). That’s a negative quantity, meaning that over the course of the whole cycle work was done on the environment. Since the energy change is zero, the heat added perfectly balances the work, so {Q}_{cycle}=({P}_{2}-{P}_{1})({V}_{2}-{V}_{1}). That’s a positive quantity. As mentioned at the end of the last Thermo Thursday, this cycle takes in heat energy and converts it to work done on the environment. So these results are as I expected.