Thermo Thursday: Compression of an Ideal Gas

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. Adiabat 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
Raising both sides of the equation to the power 2/f  and moving the constant terms to the right side gives
I can then combine the volume terms,

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
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
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.


IsoAdiaBubble 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.


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  1. In problem 1.36 part a.) there is a typo which states “and final pressure of 7 atmospheres, gives a final volume of 0.25 atmospheres.” which should be corrected by replacing “0.25 atmospheres.” to 0.25 liters. As liters is a unit of volume as atmospheres is to pressure.

    Juan A.

  2. Nice catch, thanks. Fixed.

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