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## PHY452H1S Basic Statistical Mechanics. Problem Set 4: Ideal gas

Posted by peeterjoot on March 3, 2013

# Disclaimer

## 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?

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.

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)

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.

## An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 3, 2013

That compilation now all of the following too (no further updates will be made to any of these) :

February 28, 2013 Rotation of diatomic molecules

February 28, 2013 Helmholtz free energy

February 26, 2013 Statistical and thermodynamic connection

February 24, 2013 Ideal gas

February 16, 2013 One dimensional well problem from Pathria chapter II

February 15, 2013 1D pendulum problem in phase space

February 14, 2013 Continuing review of thermodynamics

February 13, 2013 Lightning review of thermodynamics

February 11, 2013 Cartesian to spherical change of variables in 3d phase space

February 10, 2013 n SHO particle phase space volume

February 10, 2013 Change of variables in 2d phase space

February 10, 2013 Some problems from Kittel chapter 3

February 07, 2013 Midterm review, thermodynamics

February 06, 2013 Limit of unfair coin distribution, the hard way

February 05, 2013 Ideal gas and SHO phase space volume calculations

February 03, 2013 One dimensional random walk

February 02, 2013 1D SHO phase space

February 02, 2013 Application of the central limit theorem to a product of random vars

January 31, 2013 Liouville’s theorem questions on density and current

January 30, 2013 State counting

## PHY452H1S Basic Statistical Mechanics. Lecture 9: Lightning review of thermodynamics. Taught by Mr. (Eric) Kin-Ho Lee

Posted by peeterjoot on February 13, 2013

# Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

# Lightning review of thermodynamics

First law

Energy conservation.

• Work. Macroscopic control
• heat. Uncontrollable (microscopically)

This is summarized by the differential relationship

\begin{aligned}dE = d W + d Q.\end{aligned} \hspace{\stretch{1}}(1.2.1)

Examples of work

We have many types of work (in contrast to only one type of heat). Exmaples

1. $-P dV = d W$
2. $q\mathbf{E} \cdot dl$
3. $k x dx$
4. $H dm$

Homework: verify the signs of these.

We put these into a general form, to first order, of

\begin{aligned}d W_i = f_i dx_i,\end{aligned} \hspace{\stretch{1}}(1.2.2)

where we assume that higher order terms are not significant.

\begin{aligned}d W = \sum_i d W_i = \sum_i f_i dx_i.\end{aligned} \hspace{\stretch{1}}(1.2.3)

Heat

We have only one type of heat, which we loosely describe as something imbued by contact with a “hotter” system, as in (Fig 1)

Fig1: System in contact with heat source

This is defined as the condition where we have no heat exchange with the environment, or

\begin{aligned}d Q = 0.\end{aligned} \hspace{\stretch{1}}(1.2.4)

We contrast this with heating processes for which we have

\begin{aligned}d W = 0.\end{aligned} \hspace{\stretch{1}}(1.2.5)

Since we have $N$ coordinates ($d W = \sum_{i = 1}^N f_i dx_i$). We can think about an $n + 1$ dimensional space, where

1. $N$-dimensions are $x_i$
2. 1 dimension that characterizes heat exchange.

## n = 1

Given work on gas

\begin{aligned}d W = -P dV\end{aligned} \hspace{\stretch{1}}(1.2.6)

We have a coordinate, not yet precisely defined, for which fixed levels indicate that there is no heat exchange occuring, as in (Fig 2)

Fig2: Adiabatic and heat exchange processes

We’ll call this axis $\sigma$, the thermodynamic entropy.

We’ve been introduced to statistical entropy

\begin{aligned}S = k_B \ln \Omega.\end{aligned} \hspace{\stretch{1}}(1.2.7)

We’ll assume for now that these are not related and will eventually figure out the connection between these two concepts.

## n = 2

A representation of a adiabatic, or constant $\sigma$-hypersurface process is given in (Fig 3), a heating/cooling process with transition between $\sigma$-hypersurfaces in (Fig 4), and a cyclic process, in (Fig 5)

Fig4: Heat exchange process

Fig 5: Cyclic process

The cyclic process is one for which $dE = d W + d Q = 0$, however, this does not imply $d W = 0$ and $d Q = 0$ since we only require that the sum of the two is zero. In this whole process, we can have for example a net change in heat. Example: the engine of a car. Work is done, and heat is generated, but a car that was initially stopped and returns to its final destination, stops and cools down again, has still had significant internal action in the process.

Reversible processes

What do we mean by reversible? We mean that any of the changes in the system have been done so slowly that we could reverse the direction of the processes at any point, and should we do so, both the system and the environment will be returned to its initial state. This is an idealization that is, most of the time, a good approximation, but gives us an excellent idea of the limits of what we can theoretically describe.

Question: Why does the speed of the process make a difference?

If we are making changes to the system quickly, imagine that we are compressing a gas as in (Fig 6)

Fig 6: Fast gas compression by a piston

Doing work slowly means that the whole system can react to the change imposed. If we compressed the gas quickly, then changes to the system start only at the contact point with the piston. This can’t be reversed. If we pull the piston out at this point, none of the non-front gas particles will be able to react. The system will not be in thermal equalibrium for fast changes.