The figure below shows an apparatus for the determination of...

The figure below shows an apparatus for the determination of $C_p / C_v$ for a gas, according to the method of Clement and Desormes. A bottle $G$, of reasonable capacity (say a few litres), is fitted with a tap $H$, and a manometer $M$. The difference in pressure between the inside and the outside can thus be determined by observation of the difference $h$ in heights of the two columns in the manometer. The bottle is filled with the gas to be investigated, at a very slight excess pressure over the outside atmospheric pressure. The bottle is left in peace (with the tap closed) until the temperature of the gas in the bottle is the same as the outside temperature in the room. Let the reading of the manometer be $h_i$. The tap $H$ is then opened for a very short time, just sufficient for the internal pressure to become equal to the atmospheric pressure (in which case the manometer reads $h=0$ ). With the tap closed the bottle is left in peace for a while, until the inside temperature has become equal to the outside temperature. Let the final reading of the manometer be $h$. From the values of $h_i$ and $h_f$ it is possible to find $C_p / C_v$. (a) Derive an expression for $C_p / C_v$ in terms of $h_i$ and $h_f$ in the above experiment. (b) Suppose that the gas in question is oxygen. What is your theoretical prediction for $C_p / C_v$ at $20^{\circ} \mathrm{C}$, within the framework of statistical mechanics? FIGURE CAN'T COPY.

Key Concepts

Adiabatic Process

An adiabatic process is one in which a system does not exchange heat with its surroundings. When a gas expands or compresses rapidly, such as in the momentary opening of a tap, the temperature change that occurs is solely due to the work done by or on the gas without any heat transfer, which is essential for analyzing the sudden pressure drop in the experiment.

Isothermal Process

An isothermal process is characterized by a constant temperature throughout the transformation, typically achieved through sufficient time for heat exchange with the surroundings. In experimental setups where the gas eventually returns to equilibrium with ambient temperature, this process allows the pressure to readjust, and it is critical for interpreting the final pressure measurements.

Specific Heat Capacities and Their Ratio (? = Cp/Cv)

Specific heat capacities at constant pressure (Cp) and constant volume (Cv) quantify the amount of heat required to change the temperature of a gas under different constraints. Their ratio, ?, is a fundamental parameter that describes the thermodynamic behavior of a gas during adiabatic processes. This ratio is crucial for relating measurable quantities like pressure differences in the experiment to the intrinsic properties of the gas.

Ideal Gas Law

The ideal gas law, which relates pressure, volume, and temperature through the equation PV = nRT, provides the foundational framework for understanding the behavior of gases under various thermodynamic processes. It is instrumental in deriving relationships in experiments like the Clement and Desormes method, where pressure changes measured by a manometer are used to infer changes in the state of the gas.

Transcript

00:01 So this problem, problem 5 .14, is all about the difference between the specific heat at constant pressure, c -s -p, and the specific heat at constant volume, c -s -s -a -v.

00:14 In order to derive this relationship or this difference, let's start out with the entropy.

00:20 So part a.

00:23 Part a wants us to expand the entropy in terms of its partial derivatives.

00:29 In order to do this, you want to explicitly write out what the independent variable is.

00:33 For the entropy actually are.

00:36 S, intrepete, is a function of both temperature and volume.

00:46 I now use partial derivatives in the chain rule, and we see that when we expand, we get ds equals del s, del t, a constant v, multiplied by d t, plus del s, del s, del v, multiplied by d t, plus del s, del s, del v, a constant t.

01:16 Multiplied by dv and this according to problem 3 .33 is exactly cv divided by t right here so we could rearrange all of this and say cv over t plus del s del v constant t multiplied by dv okay so this is looking pretty good we have c's v but now we need to get an expression for c sub p.

01:58 So what we're going to do is expand dv in terms of its partial derivatives.

02:05 So here we go.

02:09 V equals v of t and p.

02:17 So using partial derivatives and the chain rule again when i expand dv i get del v, del t a constant p multiplied by d t plus del v del p at constant t multiplied by dp.

02:46 This problem, part b, and the problem also instructs us to set dp equal to zero, or pressure equals zero here.

03:02 D .p.

03:03 Equals zero.

03:06 So doing that, we get, upon plugging this guy into our above expression right here, right there, we get that, at ds, its constant pressure, equals del s, del t, its constant v, d t, plus del s, del s, del s, del v, multiplied by del s, oh, excuse me, v, del t, it's constant p times d t.

04:00 Okay, let's rearrange, and if we get the del s, the del s, del t a constant p equals del s del t constant v plus del s del s d s d 'n d 'n v at constant t multiplied by del v d 'l t a constant p and now what do we want to do well multiply through by t by t and then use the results of problem 3 .33 doing so so, we obtain that c -sip p equals c -s -v plus t del s, del s, del v, a constant t, multiplied by del v, del t, a constant p.

05:23 Okay, and there we go.

05:28 All right, so that's problem a done, problem b done, and now let's move on to problem c.

05:33 So problem c wants us to rewrite these partial derivatives in terms of more measure things and using the maxwell relations and problem 1 .46 to help us out.

05:45 So here we go.

05:52 The maximal relation that we will need is one that will eliminate the entropy.

06:02 And this is actually a good thing in real life because the entropy can be very difficult to measure.

06:07 And so we want to completely eliminate the entropy from that right -hand side.

06:12 The maximal relation that will do this for us is the one from the hemholt's free energy.

06:21 Here we go.

06:23 Del s, del v, a constant t equals del p, del t, a constant v.

06:35 And so this is the maximal relation that we are going to use.

06:39 If we then plug this in to the above expression to eliminate this entropy term, we obtain c -sip p minus c sub v equals t multiplied by d p d t a constant v times d t a constant p and finally using the results of problem 1 .46 we obtain that c sub p minus c sub v equals negative t multiplied by d v d t a constant p a constant p squared divided by dv d t or d p a constant t what we now have to do is remember um these well -known coefficients of volume expansion and um isothermal compressibility and so um this may have been like in the beginning chapters but i will just refresh your memories as to what it is so for the coefficient of volume expansion, that's typically denoted as beta, and that is typically said to be 1 over v times dv, d t, a constant p, and the isothermal compressibility, kappa t equals negative 1 over v.

08:41 The triple equal sign just means it is known as, or that is what it is defined as.

08:47 So, negative 1 over v times dv, dp, a constant t.

08:57 So using these expressions, and these are measurable things, we put them into our above expression, and we obtain, that c -sup p minus c -s -s -a -v equals negative t -beta v squared divided by negative kappa, t times v.

09:31 Or as the book wants, t v, v, beta squared over kappa t.

09:46 And there we go.

09:51 Alright.

09:53 And now for part d, we want to do the c sub p minus c of v for an idle gas.

10:03 So, d, ideal gas, we obtain for beta.

10:16 And these are things that are in the book that you can look up.

10:19 Up, i will give them to you now.

10:22 Beta equals 1 over v times nk divided by p, and that is simply 1 over t, and k t k t equals, and again this is for an idle gas, negative 1 over v, times negative n k t over p squared or rearranging 1 over.

10:56 That means for an idle gas, plugging it into this above expression, we obtain c sub p minus c sub v equals t v divided by t squared all over 1 over p.

11:22 And that equals pv over t, and this equals n times k.

11:32 Because remember, for an idle gas, pv equals n r t, which equals n r t, which equals n.

11:39 Okay, and this is exactly an agreement with equation 1 .48, i'll send the book.

11:50 All right.

11:53 Now, for part e, we are going to see, or at least argue, that c -s -p cannot be less than c -s -s -v.

12:08 Well, beta can be negative, interestingly enough, but it is squared in that formula.

12:20 So, kappa -t can never be negative.

12:23 So in other words, so for part e, although that beta, which typically can be negative, it can be negative, can be a minus sign.

12:46 Although that it is negative, because you square it, it becomes a positive due to the squaring...

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