Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

PHY452H1S Basic Statistical Mechanics. Problem Set 4: Ideal gas

Posted by peeterjoot on March 3, 2013

[Click here for a PDF of this post with nicer formatting]

Disclaimer

This is an ungraded set of answers to the problems posed.

Question: Sackur-Tetrode entropy of an Ideal Gas

The entropy of an ideal gas is given by

\begin{aligned}S = N k_{\mathrm{B}}\left( \ln \left( \frac{V}{N} \left( \frac{4 \pi m E}{3 N h^2} \right) ^{3/2} \right) + \frac{5}{2} \right)\end{aligned} \hspace{\stretch{1}}(1.1.1)

Find the temperature of this gas via (\partial S/ \partial E)_{V,N} = 1/T. Find the energy per particle at which the entropy becomes negative. Is there any meaning to this temperature?

Answer

Taking derivatives we find

\begin{aligned}\frac{1}{{T}} &= \frac{\partial {}}{\partial {E}}\left( \not{{ N k_{\mathrm{B}} \ln \frac{V}{N} }} + N k_{\mathrm{B}} \frac{3}{2} \ln \left( \frac{4 \pi m E}{3 N h^2} \right) + \not{{N k_{\mathrm{B}} \frac{5}{2} }} \right) \\ &= \frac{3}{2} N k_{\mathrm{B}} \frac{1}{{E}}\end{aligned} \hspace{\stretch{1}}(1.1.2)

or

\begin{aligned}\boxed{T = \frac{2}{3} \frac{E}{N k_{\mathrm{B}} }}\end{aligned} \hspace{\stretch{1}}(1.1.3)

The energies for which the entropy is negative are given by

\begin{aligned}\left( \frac{4 \pi m E}{3 N h^2} \right)^{3/2}\le \frac{N}{V} e^{-5/2},\end{aligned} \hspace{\stretch{1}}(1.1.4)

or

\begin{aligned}E &\le \frac{3 N h^2}{4 \pi m} \left( \frac{N}{V e^{5/2}} \right)^{2/3} \\ &= \frac{3 h^2 N^{5/3}}{4 \pi m V^{2/3} e^{5/2}}.\end{aligned} \hspace{\stretch{1}}(1.1.5)

In terms of the temperature T this negative entropy condition is given by

\begin{aligned}\not{{\frac{3 N}{2}}} k_{\mathrm{B}} T \le \not{{\frac{3 N}{2}}} \left( \frac{ N}{V} \right)^{2/3} \frac{h^2}{e^{5/2}},\end{aligned} \hspace{\stretch{1}}(1.1.6)

or

\begin{aligned}\boxed{\frac{\sqrt{2 \pi m k_{\mathrm{B}} T}}{h} \le \left( \frac{N}{V} \right)^{1/3} \frac{1}{{e^{5/4}}}.}\end{aligned} \hspace{\stretch{1}}(1.1.7)

There will be a particle density V/N for which this distance h/\sqrt{2 \pi m k_{\mathrm{B}} T} will start approaching the distance between atoms. This distance constrains the validity of the ideal gas law entropy equation. Putting this quantity back into the entropy eq. 1.1.1 we have

\begin{aligned}\frac{S}{N k_{\mathrm{B}}} = \ln \frac{V}{N} \left( \frac{\sqrt{2 \pi m k_{\mathrm{B}} T}}{h} \right)^3 + \frac{5}{2}\end{aligned} \hspace{\stretch{1}}(1.1.8)

We see that a positive entropy requirement puts a bound on this distance (as a function of temperature) since we must also have

\begin{aligned}\frac{h}{\sqrt{2 \pi m k_{\mathrm{B}} T}} \ll \left( \frac{V}{N} \right)^{1/3},\end{aligned} \hspace{\stretch{1}}(1.1.9)

for the gas to be in the classical domain. I’d actually expect a gas to liquefy before this transition point, making such a low temperature nonphysical. To get a feel for whether this is likely the case, we should expect that the logarithm argument to be

\begin{aligned}\frac{V}{N} \left( \frac{\sqrt{2 \pi m k_{\mathrm{B}} T}}{h} \right)^3\end{aligned} \hspace{\stretch{1}}(1.1.10)

at the point where gasses liquefy (at which point we assume the ideal gas law is no longer accurate) to be well above unity. Checking this for 1 liter of a gas with 10^23 atoms for hydrogen, helium, and neon respectively we find the values for eq. 1.1.10 are

\begin{aligned}173.682, 130.462, 23993.\end{aligned} \hspace{\stretch{1}}(1.1.11)

At least for these first few cases we see that the ideal gas law has lost its meaning well before the temperatures below which the entropy would become negative.

Question: Ideal gas thermodynamics

An ideal gas starts at (V_0, P_0) in the pressure-volume diagram (x-axis = V, y-axis = P), then moves at constant pressure to a larger volume (V_1, P_0), then moves to a larger pressure at constant volume to (V_1, P_1), and finally returns to (V_0, P_0), thus undergoing a cyclic process (forming a triangle in P-V plane). For each step, find the work done on the gas, the change in energy content, and heat added to the gas. Find the total work/energy/heat change over the entire cycle.

Answer

Our process is illustrated in fig. 1.1.

Fig 1.1: Cyclic pressure volume process

Step 1
This problem is somewhat underspecified. From the ideal gas law, regardless of how the gas got from the initial to the final states, we have

\begin{aligned}P_0 V_0 = N_0 k_{\mathrm{B}} T_0\end{aligned} \hspace{\stretch{1}}(1.0.12a)

\begin{aligned}P_0 V_1 = N_1 k_{\mathrm{B}} T_1\end{aligned} \hspace{\stretch{1}}(1.0.12b)

So a volume increase with fixed P implies that there is a corresponding increase in N T. We could have for example, an increase in the number of particles, as in the evaporation process illustrated of fig. 1.2, where a piston held down by (fixed) atmospheric pressure is pushed up as the additional gas boils off.

Fig 1.2: Evaporation process under (fixed) atmospheric pressure

Alternately, we could have a system such as that of fig. 1.3, with a fixed amount of gas is in contact with a heat source that supplies the energy required to induce the required increase in temperature.

Fig 1.3: Gas of fixed mass absorbing heat

Regardless of the source of the energy that accounts for the increase in volume the work done on the gas (a negation of the positive work the gas is performing on the system, perhaps a piston as in the picture) is

\begin{aligned}d W_1 = - \int_{V_0}^{V_1} p dV = -P_0 (V_1 - V_0).\end{aligned} \hspace{\stretch{1}}(1.0.13)

Let’s now assume that we have the second sort of configuration above, where the total amount of gas is held fixed. From the ideal gas relations of eq. 1.0.12.12, and with \Delta V = V_1 - V_0, \Delta T = T_1 - T_0, and N_1 = N_0 = N, we have

\begin{aligned}P_0 \Delta V = N k_{\mathrm{B}} \Delta T.\end{aligned} \hspace{\stretch{1}}(1.0.14)

The change in energy of the ideal gas, assuming three degrees of freedom, is

\begin{aligned}d U = \frac{3}{2} N k_{\mathrm{B}} \Delta T = \frac{3}{2} P_0 \Delta V.\end{aligned} \hspace{\stretch{1}}(1.0.15)

The energy balance then requires that the total heat absorbed by the gas must include that portion that has done work on the system, plus the excess kinetic energy of the gas. That is

\begin{aligned}d Q_1 &= \frac{3}{2} P_0 \Delta V + P_0 \Delta V \\ &= \frac{5}{2} P_0 \Delta V.\end{aligned} \hspace{\stretch{1}}(1.0.16)

Step 2

For this leg of the cycle we have no work done on the gas

\begin{aligned}d W_2 = -\int_{V_1}^{V_1} P dV = 0.\end{aligned} \hspace{\stretch{1}}(1.0.17)

We do, however have a change in energy. The energy of the gas is

\begin{aligned}U = \frac{3}{2} N k_b T = \frac{3}{2} P V.\end{aligned} \hspace{\stretch{1}}(1.0.18)

With \Delta P = P_1 - P_0, the change of energy of the gas, the total heat absorbed by the gas, is

\begin{aligned}dU_2 = d Q_2 = \frac{3}{2} V_1 \Delta P.\end{aligned} \hspace{\stretch{1}}(1.0.19)

Step 3

For the final leg of the cycle, the work done on the gas is

\begin{aligned}d W_3 &= -\int_{V_1}^{V_0} P dV \\ &= \int_{V_0}^{V_1} P dV \\ &= \Delta V \frac{P_0 + P_1}{2}.\end{aligned} \hspace{\stretch{1}}(1.0.20)

This is positive this time
Unlike the first part of the cycle, the work done on the gas is positive this time (work is being done on the gas to both compress it). The change in energy of the gas, however, is negative, with the difference between final and initial energy being

\begin{aligned}dU_3 &= \frac{3}{2} (P_0 V_0 - P_1 V_1)  \\ &= -\frac{3}{2} (P_1 V_1 - P_0 V_0) <0.\end{aligned} \hspace{\stretch{1}}(1.0.21)

The simultaneous compression and the pressure reduction require energy to be removed from the gas. We must have a negative change in heat d Q < 0, with heat emitted in this phase of the cycle. This can be verified explicitly

\begin{aligned}d Q_3 &= dU - d W \\ &= -\frac{3}{2} (P_1 V_1 - P_0 V_0) - \frac{1}{{2}} \Delta V (P_1 + P_0)< 0.\end{aligned} \hspace{\stretch{1}}(1.0.22)

Changes over the complete cycle.

Summarizing the results from each of the phases, we have

\begin{aligned}d W_1 = -P_0 \Delta V\end{aligned} \hspace{\stretch{1}}(1.0.23a)

\begin{aligned}d Q_1 = \frac{5}{2} P_0 \Delta V \end{aligned} \hspace{\stretch{1}}(1.0.23b)

\begin{aligned}d U_1 = \frac{3}{2} P_0 \Delta V \end{aligned} \hspace{\stretch{1}}(1.0.23c)

\begin{aligned}d W_2 = 0 \end{aligned} \hspace{\stretch{1}}(1.0.24a)

\begin{aligned}d Q_2 = \frac{3}{2} V_1 \Delta P \end{aligned} \hspace{\stretch{1}}(1.0.24b)

\begin{aligned}d U_2 = \frac{3}{2} V_1 \Delta P \end{aligned} \hspace{\stretch{1}}(1.0.24c)

\begin{aligned}d W_3 = \Delta V \frac{P_0 + P_1}{2} \end{aligned} \hspace{\stretch{1}}(1.0.25a)

\begin{aligned}d Q_3 = -\frac{1}{{2}} ( 3(P_1 V_1 - P_0 V_0) + \Delta V (P_1 + P_0)) \end{aligned} \hspace{\stretch{1}}(1.0.25b)

\begin{aligned}d U_3 = -\frac{3}{2} ( P_1 V_1 - P_0 V_0 )\end{aligned} \hspace{\stretch{1}}(1.0.25c)

Summing the changes in the work we have

\begin{aligned}\sum_{i = 1}^3 d W_i = \frac{1}{{2}} \Delta V \Delta P > 0.\end{aligned} \hspace{\stretch{1}}(1.0.26)

This is the area of the triangle, as expected. Since it is positive, there is net work done on the gas.

We expect the energy changes to sum to zero, and this can be verified explicitly finding

\begin{aligned}\sum_{i = 1}^3 d U_i &= \frac{3}{2} P_0 \Delta V -\frac{3}{2} ( P_1 V_1 - P_0 V_0 ) \\ &= 0.\end{aligned} \hspace{\stretch{1}}(1.0.27)

With net work done on the gas and no change in energy, there should be no net heat absorption by the gas, with a total change in heat that should equal, in amplitude, the total work done on the gas. This is confirmed by summation

\begin{aligned}\sum_{i = 1}^3 d Q_i &= \frac{5}{2} P_0 \Delta V +\frac{3}{2} V_1 \Delta P -\frac{1}{{2}} ( 3(P_1 V_1 - P_0 V_0) + \Delta V (P_1 + P_0)) \\ &= -\frac{1}{{2}} \Delta P \Delta V.\end{aligned} \hspace{\stretch{1}}(1.0.28)

Question: Adiabatic process for an Ideal Gas

Show that when an ideal monoatomic gas expands adiabatically, the temperature and pressure are related by

\begin{aligned}\frac{dT}{dP}=\frac{2}{5}\frac{T}{P}\end{aligned} \hspace{\stretch{1}}(1.0.29)

Answer

From (3.34b) of [1], we find that the Adiabatic condition can be expressed algebraically as

\begin{aligned}0 = d Q = T dS = dU + P dV.\end{aligned} \hspace{\stretch{1}}(1.0.30)

With

\begin{aligned}U = \frac{3}{2} N k_{\mathrm{B}} T = \frac{3}{2} P V,\end{aligned} \hspace{\stretch{1}}(1.0.31)

this is

\begin{aligned}0 &= \frac{3}{2} V dP + \frac{3}{2} P dV + P dV \\ &= \frac{3}{2} V dP + \frac{5}{2} P dV.\end{aligned} \hspace{\stretch{1}}(1.0.32)

Dividing through by P V, this becomes a perfect differential, and we can integrate

\begin{aligned}0 &= 3 \int \frac{dP }{P}+ 5 \int \frac{dV}{V} \\ &= 3 \ln P + 5 \ln V + \ln C \\ &= 3 \ln PV + 2 \ln V + \ln C \\ &= \ln (N k_{\mathrm{B}} T)^3 + \ln \left( \frac{N k_{\mathrm{B}} T}{P} \right)^2 + \ln C.\end{aligned} \hspace{\stretch{1}}(1.0.33)

Exponentiating yields

\begin{aligned}T^5 = C' P^2.\end{aligned} \hspace{\stretch{1}}(1.0.34)

The desired relation follows by taking derivatives

\begin{aligned}2 C' P &= 5 T^4 \frac{dT}{dP} \\ &= 5 C' \frac{P^2}{T} \frac{dT}{dP},\end{aligned} \hspace{\stretch{1}}(1.0.35)

or

\begin{aligned}\frac{dT}{dP} =\frac{2}{5} \frac{T}{P},\end{aligned} \hspace{\stretch{1}}(1.0.36)

as desired.

References

[1] C. Kittel and H. Kroemer. Thermal physics. WH Freeman, 1980.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

 
%d bloggers like this: