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#### HusamEddin

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##### thermodynamic's first law
« on: November 04, 2012, 01:48:34 PM »
Hi there,

I have some simple questions in thermodynamics ... Hope I can find the correct answers to them

*is it an irreversible process or a reversible one if the heat moves from the cold part to the hot part ?
*and is it an irreversible process or a reversible one if the heat moves from the hot part to the cold part ?

*the quantities: pressure(P) , volume(V) , concentration , are called state function ... is this sentence correct or incorrect ?

I know previously that ΔE = q - w . and ΔH = ΔE + w
*in Adiabatic processes ,, Q=0 ... so ΔE = -w ,, and ΔH = 0 ... is this right
*in Isolated System ,, Q=0 , w=0 ... so ΔE = 0 ,,, ΔH = 0 ,,, is this right
the problem is that I heared before that in Isolated system ΔE=ΔH & doesn't equal zero !!! is this really right ?
*In Isothermal processes , what is the relationship between ΔE & ΔH ,, and does any one of them qeual zero ??

*delete me*

#### HusamEddin

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##### Re: thermodynamic's first law
« Reply #1 on: November 04, 2012, 04:15:59 PM »
I obtained these from a youtube video:
in adiabatic process ... ΔE = -w
Isochoric ... ΔE = Q
Isobaric ... ΔE = Q - w
Isothermal ... ΔE = 0

but I didn't get the enthalpy changes in these processes
but I think enthalpy would be like this:

in adiabatic process ... ΔH = 0
Isochoric ... ΔH = Q ... because in isochoric ΔV=0
Isobaric ... ΔH = Q ... because in Isobaric Pressure is constant
Isothermal ... ΔH = ΔPV

are these correct just like that ?

another need of help ...
what is ΔE & ΔH in these:
*isenthalpic process
*isentropic process
*Isolated system
*open system
*close system

and lastly ... I have an important question
what is the difference between "Energy content" & "Heat content"

my regards

#### HusamEddin

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##### Re: thermodynamic's first law
« Reply #2 on: November 04, 2012, 04:19:11 PM »
and does it affect numerically the ΔE or the ΔH .. if we say reversible adiabatic process and if we say irreversible adiabatic process

#### HusamEddin

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##### Re: thermodynamic's first law
« Reply #3 on: November 05, 2012, 11:32:33 AM »

#### juanrga

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##### Re: thermodynamic's first law
« Reply #4 on: November 05, 2012, 01:43:53 PM »
Hi there,

I have some simple questions in thermodynamics ... Hope I can find the correct answers to them

*is it an irreversible process or a reversible one if the heat moves from the cold part to the hot part ?
*and is it an irreversible process or a reversible one if the heat moves from the hot part to the cold part ?

*the quantities: pressure(P) , volume(V) , concentration , are called state function ... is this sentence correct or incorrect ?

I know previously that ΔE = q - w . and ΔH = ΔE + w
*in Adiabatic processes ,, Q=0 ... so ΔE = -w ,, and ΔH = 0 ... is this right
*in Isolated System ,, Q=0 , w=0 ... so ΔE = 0 ,,, ΔH = 0 ,,, is this right
the problem is that I heared before that in Isolated system ΔE=ΔH & doesn't equal zero !!! is this really right ?
*In Isothermal processes , what is the relationship between ΔE & ΔH ,, and does any one of them qeual zero ??

*delete me*

Forum Rules: Read This Before Posting

What is an irreversible process?

Do you know some example of heat moving from the hot part to the cold part?

What is a state function?

What is an Isolated System?

What is an Isothermal process?
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#### HusamEddin

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##### Re: thermodynamic's first law
« Reply #5 on: November 06, 2012, 06:45:42 AM »
What is an irreversible process?
Irreversible means that If I transfered energy from the system to the surroundings -as an example- , but I can't return it from the surroundings to the system as it was exactly before ... a big change happened in the amount
am I wrong

Do you know some example of heat moving from the hot part to the cold part?
let's assume that I have a beam of iron ... and I heated one end of its two ends ... why not

What is a state function?
the quentity that I can take the final state of it minus the initial state of it ... and those changes happens in the middle don't mean any thing for me ...
I don't need how it happened ... and I don't need what path it took
am I wrong

Q = 0 .. no heat flow

What is an Isolated System?
no mass flow ... no heat flow ... no work

What is an Isothermal process?
no change in temperature

I answered your questions ... I think I have the basics ... but still didn't have the answers to my questions ...
waiting . and thanks
my regards

#### juanrga

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##### Re: thermodynamic's first law
« Reply #6 on: November 07, 2012, 12:49:51 PM »
What is an irreversible process?
Irreversible means that If I transfered energy from the system to the surroundings -as an example- , but I can't return it from the surroundings to the system as it was exactly before ... a big change happened in the amount
am I wrong

Ok, then consider heat flowing from the hot to the cold part of a piece of iron. Can you see the inverse process (one end of the piece gets hotter whereas the opposite end gets cooled) without changing the surrounds of the piece? No. Then the process was irreversible.

Do you know some example of heat moving from the hot part to the cold part?
let's assume that I have a beam of iron ... and I heated one end of its two ends ... why not

And experimentally you will see how the heat flows from the heated end to the other end which then becomes hotter up to that temperature is the same in all the piece. The theory will also explains why heat flows from hot to cold.

What is a state function?
the quentity that I can take the final state of it minus the initial state of it ... and those changes happens in the middle don't mean any thing for me ...
I don't need how it happened ... and I don't need what path it took
am I wrong

Ok. Then can you compute the difference in pressure, volume, concentration? Yes, because they are state functions.

Q = 0 .. no heat flow

Yes. For constant pressure ΔH = q. Therefore ΔH = 0.

What is an Isolated System?
no mass flow ... no heat flow ... no work

And the consequence is ΔE=0. I do not know where you hear the contrary.

What is an Isothermal process?
no change in temperature

Ok. Therefore you have ΔH = ΔE + pΔV + VΔp and you cannot assure that equals zero.
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#### HusamEddin

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##### Re: thermodynamic's first law
« Reply #7 on: November 07, 2012, 02:58:39 PM »
First of all ... let me thank you ... so thank you very much
I have an exam in chemistry next monday ...

OK ..
So it's like that in adiabatic process:
ΔE = -w
ΔH = 0
then ΔG = -TΔS ...
Right

in Isolated system:
ΔE = 0
ΔH = Δ(PV) <<<<  I'm not sure of this .. is it right
*by the way ... what is the correct general equation for ΔH ... is it ΔH = ΔE +Δ(PV) ... or thing else
Because I saw You wrote it like this:
ΔH = ΔE + pΔV + VΔp
!!!

in Isothermal process:
ΔE = 0
ΔH = Δ(PV) <<<<  also like that in Isolated system !!! .. I doubt .. there is something wrong !

in Isobaric process:
ΔE = Q - w
ΔH = Q
ΔG = Q - TΔS ... and for a Isobaric reversible process .. ΔG = 0 ..
Right

in Isochoric process:
ΔE = Q
ΔH = Q + VΔP ??
Right ??

please if any of my analysis was incorrect ... please tell me the correct thing

mmmmm

Finally ... It is an irreversible process if the heat moved from the cold part to the hot part ... Right ??
But my doctor told me that it is not not correct to say like this .. because this process is not found .
But I doubt ... because we can do such a process by doing some work to move heat from cold to hot .. so we can say it is an irreversible ... Am I right

my regards
« Last Edit: November 07, 2012, 03:41:38 PM by HusamEddin »

#### juanrga

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##### Re: thermodynamic's first law
« Reply #8 on: November 08, 2012, 02:44:32 PM »
First of all ... let me thank you ... so thank you very much
I have an exam in chemistry next monday ...

Good luck!

OK ..
So it's like that in adiabatic process:
ΔE = -w
ΔH = 0
then ΔG = -TΔS ...
Right

Yes in adiabatic process ΔE = -w and ΔH = 0

The definition of G for any situation is G = H - TS

which implies that ΔG = - SΔT - TΔS for adiabatic process. If, moreover, T is constant then ΔG = -TΔS

in Isolated system:
ΔE = 0
ΔH = Δ(PV) <<<<  I'm not sure of this .. is it right
*by the way ... what is the correct general equation for ΔH ... is it ΔH = ΔE +Δ(PV) ... or thing else
Because I saw You wrote it like this:
ΔH = ΔE + pΔV + VΔp
!!!

Yes, you are right. The definition of H for any situation is H = E + PV and then ΔH = ΔE + Δ(PV)

in Isothermal process:
ΔE = 0
ΔH = Δ(PV) <<<<  also like that in Isolated system !!! .. I doubt .. there is something wrong !

Isothermal means constant temperature. ΔE can be zero or not.

in Isobaric process:
ΔE = Q - w
ΔH = Q
ΔG = Q - TΔS ... and for a Isobaric reversible process .. ΔG = 0 ..
Right

You are lacking the -SΔT term in ΔG. Why do you think that ΔG = 0 for isobaric reversible process?

in Isochoric process:
ΔE = Q
ΔH = Q + VΔP ??
Right ??

Ok.

Finally ... It is an irreversible process if the heat moved from the cold part to the hot part ... Right ??
But my doctor told me that it is not not correct to say like this .. because this process is not found .
But I doubt ... because we can do such a process by doing some work to move heat from cold to hot .. so we can say it is an irreversible ... Am I right

As said above heat does not flow from the cold part to the hot part. He is correct.
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#### HusamEddin

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##### Re: thermodynamic's first law
« Reply #9 on: November 10, 2012, 07:23:40 AM »
sorry but things came together and made me crazy
look:
ΔE = -w = -PΔV  equation(1)
ΔH doesn't equal zero ... look:
H = E + PV
ΔH = ΔE + Δ(PV)
ΔH = ΔE + PΔV + VΔP
from equation(1)
ΔH = -PΔV + PΔV + VΔP
then finally:
ΔH = VΔP
and this means ΔH doesn't equal zero but equals VΔP
confused
do you have an explanation for this Contradiction

#### HusamEddin

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##### Re: thermodynamic's first law
« Reply #10 on: November 10, 2012, 07:42:10 AM »
there was a "True or False" question in an old exam tells:
Heat content is equal to energy content at constant volume (  )
and the correct answer was "True"
This is confusing ... look:

constant volume means it is an Isochoric Process:
Isochoric Process:
ΔE = Q , because ΔV = 0 then -PΔV = 0
ΔH = ΔE + Δ(PV) = ΔE + PΔV + VΔP = ΔE + VΔP ... because PΔV = 0

Energy content  = ΔE = Q
Heat content = ΔH = ΔE + VΔP
then Heat content doesn't equal Energy content at constant volume !!!!

can you help me explaining this please

another question .. also true or false
Gibb's free energy is negative for a spontaneous reversible process (   )
and the correct answer was "True" !!
also this is confusing !!
look
ΔG = ΔH - TΔS - SΔT
I can't do any thing with this equation ... I mean I can't determine the final sign if it is positive or negative for a reversible process , because ΔH here could be zero or positive or negative depending on the type of the process ... and and there is no mentioned type in the sentence !

mmmmmmmm !!!

#### juanrga

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##### Re: thermodynamic's first law
« Reply #11 on: November 10, 2012, 11:52:23 AM »
sorry but things came together and made me crazy
look:
ΔE = -w = -PΔV  equation(1)

This equation is only valid for constant pressure.

ΔH doesn't equal zero ... look:
H = E + PV
ΔH = ΔE + Δ(PV)
ΔH = ΔE + PΔV + VΔP
from equation(1)
ΔH = -PΔV + PΔV + VΔP
then finally:
ΔH = VΔP
and this means ΔH doesn't equal zero but equals VΔP
confused
do you have an explanation for this Contradiction

You assumed constant pressure, therefore ΔH = VΔP = 0.
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#### juanrga

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##### Re: thermodynamic's first law
« Reply #12 on: November 10, 2012, 12:12:50 PM »
there was a "True or False" question in an old exam tells:
Heat content is equal to energy content at constant volume (  )
and the correct answer was "True"
This is confusing ... look:

constant volume means it is an Isochoric Process:
Isochoric Process:
ΔE = Q , because ΔV = 0 then -PΔV = 0

Therefore heat (Q) equals energy (ΔE) at constant volume.

ΔH = ΔE + Δ(PV) = ΔE + PΔV + VΔP = ΔE + VΔP ... because PΔV = 0

Energy content  = ΔE = Q
Heat content = ΔH = ΔE + VΔP

ΔH is change in enthalpy. As said before, change in enthalpy equals heat at constant pressure and the last expression is consistent because ΔH = ΔE + VΔP = Q when ΔP=0.

another question .. also true or false
Gibb's free energy is negative for a spontaneous reversible process (   )
and the correct answer was "True" !!
also this is confusing !!
look
ΔG = ΔH - TΔS - SΔT
I can't do any thing with this equation ... I mean I can't determine the final sign if it is positive or negative for a reversible process , because ΔH here could be zero or positive or negative depending on the type of the process ... and and there is no mentioned type in the sentence !

mmmmmmmm !!!

Substituting ΔH by the value that you computed above ΔH = ΔE + PΔV + VΔP we obtain

ΔG = (ΔE + PΔV + VΔP) - TΔS - SΔT

Using the first law

ΔG = (Q + VΔP) - TΔS - SΔT

Using the second law TΔS ≥ Q

ΔG ≤ VΔP - SΔT

The Gibbs energy is used for processes at constant pressure and constant temperature. For those processes

ΔG ≤ 0
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#### HusamEddin

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##### Re: thermodynamic's first law
« Reply #13 on: November 13, 2012, 09:57:22 AM »
thanks alot my friend
my exam travelled to the next Monday
mmmmm
some thing wrong happened
while I was turning my pages (my chemistry notebook)
I found that my doctor told us that ΔE = Qrev. - w
where Qrev. means Q(reversible)
my knowledge that (q) here is not necessery to be reversible !!!