# 查看完整版本 : (concept)有關first law of thermodynamics and internal energy

## (concept)有關first law of thermodynamics and internal energy

[size=4][color=#333333][font=arial, helvetica, clean, sans-serif]A sample of an ideal gas goes through the process shown below. From A to B, the process is adiabatic; from B to C, it is isobaric with 345 kJ of energy entering the system by heat. From C to D, the process is isothermal; and from D to A, it is i[/font][/color][color=#333333][font=arial, helvetica, clean, sans-serif]sobaric with 371 kJ of energy leaving the system by heat. Determine the difference in internal energy, Eint,B − Eint,A.  (其實仲有好多題都係e一幅圖,但Q數值不同)[/font][/color][/size][size=4][color=#333333][font=arial, helvetica, clean, sans-serif]                                      [img]http://www.webassign.net/pse/p20-32.gif[/img][/font][/color]
[/size]
[size=4][color=#333333][font=arial, helvetica, clean, sans-serif]我的問題係點解唔可以用直接用E(int)=3/2 nRT=3/2 PV  or 5/2 nRT or Cv nRT 去計 [/font][/color][font=arial, helvetica, clean, sans-serif]Eint,B − Eint,A ,即係(3/2(PbVb-PaVa), [/font][/size]
[size=4][font=arial, helvetica, clean, sans-serif]delt U = Q +W  delt U =Uf-Ui [/font][/size]
[align=left][size=4][font=arial, helvetica, clean, sans-serif]而要用[/font][font=arial, helvetica, clean, sans-serif][color=#333333]first law of thermodynamics(我一開始係用first law去計 但發現[/color][/font][font=arial, helvetica, clean, sans-serif]點解唔直接用E(int)計 ,試下之後答案不同)[/font][/size][/align]
[size=4][font=arial, helvetica, clean, sans-serif][color=#333333]求高人指點[/color][/font]:smile_42::smile_44:[/size]

↗亢龍無悔↙ 2012-11-16 02:50 PM

isobaric process的話,因為pressure係constant,將個p抽出個integration即係搵長方形面積
isovolumetric process的話,area = 0,即係delta W = 0
isothermal process的話,都要有條equation做integration

[[i] 本帖最後由 ↗亢龍無悔↙ 於 2012-11-16 03:12 PM 編輯 [/i]]

[quote]原帖由 [i]↗亢龍無悔↙[/i] 於 2012-11-16 02:50 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=347560993&ptid=21135165][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

Ub=3/2PbVb
Ua=3/2PaVa

isothermal process的話,其實係delt U =0 ,as U=3/2nRT,T不變 ( 唔洗計workdone)
adiabatic process, Q=0 ,delt U = W, W要integration, PV^(Y)=constant
isobaric , delt U= Q-P delt V

[[i] 本帖最後由 問題小朋友 於 2012-11-16 07:52 PM 編輯 [/i]]

Zzlaz 2012-11-17 07:25 PM

[url=http://www.discuss.com.hk/android][img=100,23]http://i.discuss.com.hk/d/images/r10/androidD.jpg[/img][/url]

[quote]原帖由 [i]Zzlaz[/i] 於 2012-11-17 07:25 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=347645564&ptid=21135165][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

[img]http://i.discuss.com.hk/d/images/r10/androidD.jpg[/img] [/quote]
n係無寫過係咩,但nRT=PV,不是嗎?

Zzlaz 2012-11-17 11:57 PM

↗亢龍無悔↙ 2012-11-19 12:45 PM

..

[[i] 本帖最後由 問題小朋友 於 2012-11-21 09:57 PM 編輯 [/i]]

↗亢龍無悔↙ 2012-11-20 01:23 AM

lung0807 2012-11-20 05:17 PM

where (3/2nRT) is just the total K.E. of molecule ,

jmlo 2012-11-21 01:44 PM

## 回覆 1#, 3# 的帖子

Here are some thermodynamic concepts you have to make clear. I presume you are at university level (right?), so I try to elaborate the ideas in a more technical way.

[quote] 我的問題係點解唔可以用直接用E(int)=3/2 nRT=3/2 PV  or 5/2 nRT or Cv nRT 去計 Eint,B − Eint,A ,即係(3/2(PbVb-PaVa),
delt U = Q +W  delt U =Uf-Ui

Recall that internal energy is a state function depending only on temperature; therefore, the choice of path is irrelevant. Calculations for A ---> B and for A ---> D ---> C ---> B should give the same answer if they are done correctly. Usually, people do the calculations using the path which can be most easily solved.

[quote] 而delt U = Q + W  ,其中U 係internal energy,  U 包括 P.E and K.E ,但 in ideal gas , assume P.E. =0

Ub=3/2PbVb
Ua=3/2PaVa  [/quote]
There are two mistakes. Firstly, since we are almost always dealing with ideal gases (or at least assume that ideal gas behavior is approximately valid at all times; this can be kind of achieved by controlling the pressure and temperature), we never really consider interatomic or intermolecular force. On the other hand, apart from kinetic energy, internal energy includes also any other form of intramolecular energy contributions. Note that molecules can both vibrate and rotate; therefore, there will be energy stored in form of vibrational and rotational motions, and they all contribute to the overall internal energy of the system.

Secondly, the formula U = 3/2 nRT is only valid for monatomic species which possess only translational motions! Going beyond that, you have to consider each degree of freedom of atoms in the molecule plus other types of motions (as described above). Hence, if you use U = 3/2 nRT directly in calculations you will not get the right answer.

FYI: This is indeed a common misunderstanding for AL or even undergraduate students since the teachers do not explain clearly enough how this equation is obtained in the first place and what kinds of assumptions are employed.

[quote] 我一開始係用first law去計 但發現點解唔直接用E(int)計 ,試下之後答案不同 [/quote]
Since U = 3/2 nRT is not the right equation to be used, certainly the answers are different.

[quote] 其實另一個想法只純粹考慮佢係B 和 A 的U值, as PV=nRT ,so U=3/2nRT=3/2PV Ui+deltU=Uf  ,那Uf-Ui=deltU=3/2PfVf-3/2PiVi

The origin of problem is the fact that you don't even know how the internal energy of the system at point A and point B can be computed. Your idea is making perfect sense if these values are given or known. Otherwise, integration along the path is the only way you can find out the change of U from A to B.

[quote] 要轉做pa [/quote]
You can perform the calculations using other units of pressure as long as appropriate units for other variables are used.

[quote] n係無寫過係咩,但nRT=PV,不是嗎? [/quote]
This is just the ideal gas law. However, nothing about the gas is indeed mentioned in the question. You cannot assume any property of it.

jmlo 2012-11-21 01:49 PM

## 回覆 2# 的帖子

[quote] 佢地最鐘意出一個cycle,包哂4種process(至少3種),因為各自唔同嘅process,各自會有delta U, delta Q或者delta W會係0

Indeed, the reason why a thermodynamic cycle is usually asked is that the change of internal energy cannot be calculated directly, and a secondary path must be used instead. In this example, since calculation making use of A ---> B is not possible, one must create a new path (i.e., A ---> D  ----> C  ---> B) that connects the initial and final points.

jmlo 2012-11-21 01:52 PM

## 回覆 7# 的帖子

You can find detailed discussions and sample calculations for this kind of problem in standard physical chemistry textbooks (e.g. Mortimer or Atkins) or chemical thermodynamics books (e.g. Smith).

jmlo 2012-11-21 01:56 PM

## 回覆 10# 的帖子

[quote] 應該係因為internal not only depend on temperature , but also potential energy,
where (3/2nRT) is just the total K.E. of molecule , [/quote]
Internal energy for an ideal gas is a state function which depends only on temperature. All energy terms (i.e., kinetic energy, rotational energy, vibrational energy) can be described in terms of temperature. Intermolecular potential energy is always assumed zero for ideal gas.

ideal gas law 唔係適用於ideal gas (question stated)? if the particles rotate or vibrate, is it valid?

[the formula U = 3/2 nRT is only valid for monatomic species which possess only translational motions!]
If it isn't the monotonic species and considering All energy terms (i.e., kinetic energy, rotational energy, vibrational energy), is the U usually bigger than 3/2 nRT ?

[[i] 本帖最後由 問題小朋友 於 2012-11-21 10:34 PM 編輯 [/i]]

jmlo 2012-11-22 12:02 AM

## 回覆 15# 的帖子

[quote] ideal gas law 唔係適用於ideal gas (question stated)? if the particles rotate or vibrate, is it valid? [/quote]
The question only mentions that the gas is ideal, but nothing about its properties has been stated. Therefore, you can't assume a monatomic ideal gas and apply U = 3/2 nRT directly. This is indeed the reason why you need to use the first law of thermodynamics to calculate U(B) - U(A) instead of computing the difference in internal energy directly.

Ideal gas law is valid for all types of gases (monatomic, diatomic, etc) as long as all basic assumptions are fulfilled (or approximately fulfilled).

[quote] If it isn't the monotonic species and considering All energy terms (i.e., kinetic energy, rotational energy, vibrational energy), is the U usually bigger than 3/2 nRT ? [/quote]
Yes. And this explains why old classical thermodynamics does not yield correct dU for processes involving polyatomic gases. The deviation gets bigger when temperature goes higher.

↗亢龍無悔↙ 2012-11-22 01:36 AM

[quote]原帖由 [i]jmlo[/i] 於 21-11-2012 01:44 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=347929614&ptid=21135165][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
There are two mistakes. Firstly, since we are almost always dealing with ideal gases (or at least assume that ideal gas behavior is approximately valid at all times; this can be kind of achieved by controlling the pressure and temperature), we never really consider interatomic or intermolecular force. On the other hand, apart from kinetic energy, [color=red]internal energy includes also any other form of intramolecular energy contributions. Note that molecules can both vibrate and rotate[/color]; therefore, there will be energy stored in form of [color=red]vibrational and rotational motions[/color], and they all contribute to the overall internal energy of the system.

Secondly, the formula U = 3/2 nRT is only valid for monatomic species which possess only translational motions! Going beyond that, you have to consider each [color=red]degree of freedom[/color] of atoms in the molecule plus other types of motions (as described above). Hence, if you use U = 3/2 nRT directly in calculations you will not get the right answer.

FYI: This is indeed a common misunderstanding for AL or even undergraduate students since the teachers do not explain clearly enough how this equation is obtained in the first place and what kinds of assumptions are employed.[/quote]

jmlo 2012-11-22 05:44 AM

## 回覆 17# 的帖子

[quote] 好西利,呢啲我係本書到見過,睇黎jmlo師兄唔止major chemistry [/quote]
I learnt these concepts in advanced physical chemistry courses (e.g. quantum chemistry, classical and statistical thermodynamics). In fact, these topics are also covered in introductory physical chemistry courses but professors usually skip the detailed explanations as basic thermodynamic concepts are already too much for chemistry students.

[quote] 我想問咩係degree of freedom?

In general sense, degree of freedom refers to the parameters on which a quantity depends. For example, given a function y = f(x), we say that y is dependent on x and it has one degree of freedom since there is only one independent variable.

This term (DOF) bears slightly different meanings when used in mathematics, physics, chemistry or engineering science. In describing the motion of molecules, we can define its overall movement in space by specifying the position of each constituent atom (i.e., its x-, y- and z-coordinates in Cartesian coordinate system). It means that each atom has three degrees of freedom. If each molecule is made of N atoms, the total DOFs the molecule possesses would be 3N.

Note that (x,y,z) is only one way of describing the motion of molecules, and this method is very often inconvenient. Therefore, we can "translate" this description into the one where the DOFs are classified as "translation", "rotation" and "vibration". The overall number of DOFs remains the same regardless of the description. Since 3 DOFs are reserved for translation and other 3 for rotations, the remaining DOFs which describe the vibrations of a given molecule are 3N-6. This is the origin of the so-called 3N-6 rule in IR spectroscopy. For linear molecules, one rotational mode is replaced by vibration.

In mechanics, DOFs describe the independent variables used to define the translation and rotation of an object. That's pretty much all I know about DOFs in mechanics as I have had zero training in this field. :loveliness:

[size=4]if from B to C it is isobaric with [/size][size=7]100 kJ[/size][size=4] of energy entering the system by heat, is it possible?[/size][size=4]1atm=1.013 X 10^5[/size]
[size=4]If using the first law of thermodynamics, delt U= 100kJ- 3 X 1.013 X 10^5 (0.4-0.09)=5791 J  from B to A[/size]
[size=4]
[/size]
[size=4][quote]If it isn't the monotonic species and considering All energy terms (i.e., kinetic energy, rotational energy, vibrational energy), is the U usually bigger than 3/2 nRT ?
Yes. And this explains why old classical thermodynamics does not yield correct dU for processes involving polyatomic gases. The deviation gets bigger when temperature goes higher.
[/quote]
[/size]
[size=4]
[/size]
[size=4]so if using 3/2 nRT, that is the minimum value for the internal energy of ideal gas[/size]
[size=4](don't know its properties, only know it is ideal gas)[/size]
[size=4]so from B to C, internal energy change >=   delt(3/2 nRT )[/size]
[size=4][quote]Ideal gas law is valid for all types of gases (monatomic, diatomic, etc) as long as all basic assumptions are fulfilled (or approximately fulfilled).[/quote][/size]
[size=4]so internal energy change >= delt (3/2 PV) =  3/2 X 3 X 1.013 X 10^5 (0.4-0.09)=1.41 X 10^5 J[/size]
[size=4]which is bigger than 5791J . That means minimum [/size]internal energy change is 1.41 X 10^5 J ,however is bigger than 5791 J
[size=4]So is it possible that [/size]from B to C it is isobaric with [size=7]100 kJ[/size] of energy entering the system by heat according to the picture?
[size=4]
[/size]
[size=4]
[/size]

jmlo 2012-11-22 11:56 PM

## 回覆 19# 的帖子

Note that 3/2 nRT is the internal energy stored in a monatomic ideal gas; for polyatomic gases the internal energy is higher because they possess more degrees of freedom (as we have discussed previously). However, in a thermodynamic cycle, not all this internal energy can be transformed into heat or work done unless one can freeze all atoms (in other words, all kinetic energy is removed). On the other hand, energy can be input into the system to increase the total internal energy until a threshold where molecules start to fall apart.

## 回覆 20# 的帖子

did you mean that it is possible?:smile_41:
did you mean that heat + work done will not equal to delt internal energy?
heat + workdone + other energy = delt internal energy?
:smile_41: :smile_41:sorry, i dont understand

jmlo 2012-11-24 02:25 AM

## 回覆 21# 的帖子

Noop. I did not mean that the first law is wrong. By conservation of energy,
dU = dq + dw
is always true. What I mean here is that while the change of internal energy, dU, is the sum of heat change and work done, the total internal energy (not the change) is the sum of these plus the other forms of energy stored in the system. This extra energy may be partly, but not all, converted to heat released to and the work done against the surrounding.

actually, heat and work done is process of transferring energy,
so did you mean the U= kinetic energy + other forms of energy?
When from B to C , kinetic energy increase, but other forms of energy decrease as extra energy may be partly converted to heat released to and the work done against the surrounding.
so d U= dq+ dW and B to C is possible.

however, is it still considered as ideal gas?

[[i] 本帖最後由 問題小朋友 於 2012-11-24 01:17 PM 編輯 [/i]]

jmlo 2012-11-24 02:38 PM

## 回覆 23# 的帖子

[quote] actually, heat and work done is process of transferring energy,
so did you mean the U= kinetic energy + other forms of energy? [/quote]
Certainly heat and work are means of changing the value of the internal energy. Perhaps my explanation may have confused you. You are right; the total internal energy U of a system is the sum of the kinetic energy and the energy that exists in other forms (e.g. vibration, rotation, electronic).

[quote] When from B to C , kinetic energy increase, but other forms of energy decrease as extra energy may be partly converted to heat released to and the work done against the surrounding.
so d U= dq+ dW and B to C is possible. [/quote]
Yes.

[quote] however, is it still considered as ideal gas? [/quote]
Note that the behaviors discussed above have nothing to do that contradicts the ideal gas assumptions.

thx a lot

However, there are some other questions, which require using 3/2nRT to get the answer but they only stated the gas is ideal gas. They confused me .

[[i] 本帖最後由 問題小朋友 於 2012-11-24 04:17 PM 編輯 [/i]]

Zzlaz 2012-11-25 02:42 AM

jmlo 2012-11-25 03:45 AM

## 回覆 25# 的帖子

I see. Chemical thermodynamics is not an easy subject really. :smile_42: