問題小朋友 2012-11-15 01:53 PM

[/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

識得係問題入面搵問題,比個叻你先:smile_o12:

因為讀physics係要咁讀,先唔會比啲出卷人fake

但係睇你嘅問法,你嘅concept有少少錯左@@

首先,你要搵workdone by gas的話,係應該搵"area under p-V graph",即係要做integration

條formula係~~~ negative integration (from final volume to initial volume) p dV

得出嘅value,就自然知道係workdone by gas,定係workdone on gas

即係話,如果你有AB條curve嘅equation的話,就的確可以直接搵到

但係AL physics係考concept唔係考maths,所以AL絕對唔會咁出法

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

出一個cycle可以一次過考哂全部concept,換轉你係出卷嘅人,你會點出呢?:smile_o10:

你所講嘅3/2(PbVb-PaVa),只不過係計緊KE difference between B and A,而唔係做緊integration!!!!!咁當然係錯

題外話,因為adiabatic process唔係一個linear process,如果要搵workdone by gas,就一定要做integration而唔可以直接用梯形面積

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

太耐無做thermodynamics,我想問係咪要將atm轉Pa先做到@@?

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

因為讀physics係要咁讀,先唔會比啲出卷人fake

但係睇你嘅問法,你嘅concept有少少錯左@@

首先,你要搵workdone by gas的話,係應該搵"area under p-V graph",即係要做integration

條formula係~~~ negative integration (from final volume to initial volume) p dV

得出嘅value,就自然知道係workdone by gas,定係workdone on gas

即係話,如果你有AB條curve嘅equation的話,就的確可以直接搵到

但係AL physics係考concept唔係考maths,所以AL絕對唔會咁出法

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

出一個cycle可以一次過考哂全部concept,換轉你係出卷嘅人,你會點出呢?:smile_o10:

你所講嘅3/2(PbVb-PaVa),只不過係計緊KE difference between B and A,而唔係做緊integration!!!!!咁當然係錯

題外話,因為adiabatic process唔係一個linear process,如果要搵workdone by gas,就一定要做integration而唔可以直接用梯形面積

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

太耐無做thermodynamics,我想問係咪要將atm轉Pa先做到@@?

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

問題小朋友 2012-11-16 07:33 PM

[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]

識得係問題入面搵問題,比個叻你先:smile_o12:

因為讀physics係要咁讀,先唔會比啲出卷人fake

但係睇你嘅問法,你嘅concept有少少錯左@@

首先,你要搵workdone by gas的話,係應該搵"area under ... [/quote]

我知要 integration計workdone , W= - integrate P dv ,即area under p-V graph

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

那U自然只係sum of k.e. ,即U=3/2nRT(for translational motion only),so

Ub=3/2PbVb

Ua=3/2PaVa

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 :smile_44:Ui+deltU=Uf ,那Uf-Ui=deltU=3/2PfVf-3/2PiVi

佢係curve, 如果要計W當然要 integration,

但我現在只考慮其實A,B 的U(3/2nRT)直接計(應該唔洗用integration,唔理workdone),兩點的internal energy 相減,那定是delt U, 那樣問題在那?:smile_41:

問題都係問delt U

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

要轉做pa

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

識得係問題入面搵問題,比個叻你先:smile_o12:

因為讀physics係要咁讀,先唔會比啲出卷人fake

但係睇你嘅問法,你嘅concept有少少錯左@@

首先,你要搵workdone by gas的話,係應該搵"area under ... [/quote]

我知要 integration計workdone , W= - integrate P dv ,即area under p-V graph

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

那U自然只係sum of k.e. ,即U=3/2nRT(for translational motion only),so

Ub=3/2PbVb

Ua=3/2PaVa

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 :smile_44:Ui+deltU=Uf ,那Uf-Ui=deltU=3/2PfVf-3/2PiVi

佢係curve, 如果要計W當然要 integration,

但我現在只考慮其實A,B 的U(3/2nRT)直接計(應該唔洗用integration,唔理workdone),兩點的internal energy 相減,那定是delt U, 那樣問題在那?:smile_41:

問題都係問delt U

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

要轉做pa

[[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

邊到寫n equals to 1??

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

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

問題小朋友 2012-11-17 10:53 PM

[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]

邊到寫n equals to 1??

[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?

邊到寫n equals to 1??

[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

相差大唔大？?

我計到delta U= 18.91 KJ

用你果條式 5/2 (PbVb-PaVa) 係delta U = 17.73 KJ

我計到delta U= 18.91 KJ

用你果條式 5/2 (PbVb-PaVa) 係delta U = 17.73 KJ

↗亢龍無悔↙ 2012-11-19 12:45 PM

查左兩本書都無講過呢個問題:smile_42:

佢只係define internal energy係咩東東.........

佢只係define internal energy係咩東東.........

問題小朋友 2012-11-19 01:55 PM

..

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

[[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

問完上黎教返我:smile_o06:

lung0807 2012-11-20 05:17 PM

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 ,

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

jmlo 2012-11-21 01:44 PM

[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

而要用first law of thermodynamics(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]

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

那U自然只係sum of k.e. ,即U=3/2nRT(for translational motion only),so

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

佢係curve, 如果要計W當然要 integration,

但我現在只考慮其實A,B 的U(3/2nRT)直接計(應該唔洗用integration,唔理workdone),兩點的internal energy 相減,那定是delt U, 那樣問題在那? [/quote]

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

出一個cycle可以一次過考哂全部concept,換轉你係出卷嘅人,你會點出呢? [/quote]

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

jmlo 2012-11-21 01:56 PM

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.

問題小朋友 2012-11-21 10:19 PM

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]]

[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

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]

好西利,呢啲我係本書到見過,睇黎jmlo師兄唔止major chemistry:smile_o06:

我想問咩係degree of freedom?

我係structural mechanics見左呢個字3年,但係都唔知係咩意思:smile_42:

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]

好西利,呢啲我係本書到見過,睇黎jmlo師兄唔止major chemistry:smile_o06:

我想問咩係degree of freedom?

我係structural mechanics見左呢個字3年,但係都唔知係咩意思:smile_42:

jmlo 2012-11-22 05:44 AM

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?

我係structural mechanics見左呢個字3年,但係都唔知係咩意思 [/quote]

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:

問題小朋友 2012-11-22 07:29 PM

[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]

[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

問題小朋友 2012-11-24 12:12 AM

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

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.

問題小朋友 2012-11-24 01:13 PM

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]]

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

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.

問題小朋友 2012-11-24 04:13 PM

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]]

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

其實我想知答案係乜, 好耐無計過...

jmlo 2012-11-25 03:45 AM

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