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Author Topic: Isothermal expansion  (Read 55248 times)

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madscientist

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Isothermal expansion
« on: February 20, 2006, 03:08:37 AM »

Hi everyone,

Lets say we are looking at an Irreversible isothermal process:

Quote, "An isothermal process is a thermodynamic process in which the temperature of the system stays constant; ?T = 0. This typically occurs when a system is in contact with an outside thermal reservoir, and the system changes slowly enough to allow it to adjust to the temperature of the reservoir. The opposite extreme in which a system exchanges no heat with its surroundings is known as an adiabatic process." end quote.

en.wikipedia.org/wiki/Isothermal

so, if an isothermal process means that Delta T =0, does that mean delta P=0 also, and if so how can the volume change.

hear is the question im working on and why im so confused by this.

A sample of 1.0 mol He (assume ideal) with a Cv = 3/2 R, initially at 298K and 10L is expanded, with the surroundings maintained at 298K, to a final volume of 20L, in two ways:

n=1mol, Cv= 12.471 JK, Tinitial = 298K, Vinitial = 10L, Vfinal = 20L

Tsurr = constant = 298K  (therefore the Tsystem should be constant yes ??? )

(a.) Isothermally and reversibly

(b.) Isothermally against a constant external pressure of 0.50atm

calculate Delta S, Delta H, Delta T, Delta A and Delta G for each path.

ok, so i found it fairly easy to answer part a, and am confident with my answer:

(a.)isothermally and reversibly

wrev=-nRT ln(Vfinal/Vinitial) = -1717.322 J

* for an isothermal process either reversible or irreversible the internal energy change, Delta U, is equal to zero (dU=0)
therefore:

q = -w = 1717.332 J

Delta S = qrev / T = 5.763 J/K

If: PiVi = PfVf = nRT

then: Delta(pV) = 0

Delta H = Delta U + Delta(pV) = 0 + 0 = 0

Delta T = 0 (isothermal)

Delta G = Delta H -T Delta S = 1717.3144 J ( which is = q ???)

Delta A = Wrev = -1717.322 J

my working for  (b.) is as follows, I get stuck at Delta T.

(b.)Isothermally against Pext = 0.50atm
     (Isothermally and Irreversibly then isnt it ???)

wirrev = -Pext . Delta V = -506.6 J

Delta U = q + w = 0,  (q = -w)

Delta H = Delta U + Delta (pV) = 0 + 0 = 0

Delta S = nR ln (Vfinal / Vinitial) = 5.763 J/K

now for Delta T, it should equal to zero being an isothermal process shouldnt it?

Or is it:

wirrev = -Pext . Delta V = Cv . Delta T

so: Delta T = wirrev / Cv = -40.623K

and for the Delta G and Delta A values, how are they calculated for an isothermally Irreversible process?

If anyone can check my work for part (a.) and the work attempted for part (b.) it would be very helpfull.

cheers,

madscientist :albert:
« Last Edit: February 20, 2006, 03:12:57 AM by madscientist »
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Donaldson Tan

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Re:Isothermal expansion
« Reply #1 on: February 20, 2006, 11:57:24 PM »

so, if an isothermal process means that Delta T =0, does that mean delta P=0 also, and if so how can the volume change.

That is only true for an isoentropic (reversible & adiabatic) process
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Re:Isothermal expansion
« Reply #2 on: February 22, 2006, 04:02:43 AM »

(a.) Isothermally and reversibly

calculate Delta S, Delta H, Delta T, Delta A and Delta G for each path.

wrev=-nRT ln(Vfinal/Vinitial) = -1717.322 J

* for an isothermal process either reversible or irreversible the internal energy change, Delta U, is equal to zero (dU=0)
therefore:

q = -w = 1717.332 J

Delta S = qrev / T = 5.763 J/K

If: PiVi = PfVf = nRT

then: Delta(pV) = 0

Delta H = Delta U + Delta(pV) = 0 + 0 = 0

Delta T = 0 (isothermal)

Delta G = Delta H -T Delta S = 1717.3144 J ( which is = q ???)

Delta A = Wrev = -1717.322 J

Yes, you are correct for part (a)
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madscientist

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Re:Isothermal expansion
« Reply #3 on: February 22, 2006, 12:32:59 PM »

Cheers geodome,

I actually stuck my nose in book and did some reading,(funny how that helps ;), I didnt realise that anything with a capital, for eg dU, dT, dH.. is a state function and as long as the final conditions are the same in a reversible process as they are in an irreversible process their values are the same, their said to be path independant, whereas lower case expressions like work and heat w and q, are path dependant and will be different values for each process.

Is this reasoning correct?

cheers,

madscientist :albert:
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Re:Isothermal expansion
« Reply #4 on: February 22, 2006, 03:26:30 PM »

state functions are function whose values are path-independent.

the notation for capital is not for highlighting state functions. small capital serves to highlight the value of the state function per unit mass

eg. we use h or H to represent enthalpy. however, the unit for h is kJ/kg whereas the unit for H is kJ.

so when u see w, it actually refers to the amount of work done by the system per unit mole of substance in the system. whereas W refer to the amount work done by the system.

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madscientist

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Re:Isothermal expansion
« Reply #5 on: February 22, 2006, 08:25:17 PM »

Thats extremely handy to know geodome, Thanks heaps for all your help.

cheers,

madscientist
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Re:Isothermal expansion
« Reply #6 on: February 23, 2006, 01:29:12 PM »

Or is it:

wirrev = -Pext . Delta V = Cv . Delta T

so: Delta T = wirrev / Cv = -40.623K

and for the Delta G and Delta A values, how are they calculated for an isothermally Irreversible process?
We are considering an isothermal process, so this approach is wrong.

(b.) Isothermally against a constant external pressure of 0.50atm

calculate Delta S, Delta H, Delta T, Delta A and Delta G for each path.

(b.)Isothermally against Pext = 0.50atm
     (Isothermally and Irreversibly then isnt it ???)

wirrev = -Pext . Delta V = -506.6 J

Delta U = q + w = 0,  (q = -w)

Delta H = Delta U + Delta (pV) = 0 + 0 = 0

Delta S = nR ln (Vfinal / Vinitial) = 5.763 J/K

now for Delta T, it should equal to zero being an isothermal process shouldnt it?
We are considering the case for an isothermal & irreversible process.
dU = Q + W
since dU = n.Cv.dT where dT = 0, then dU = 0 => Q = -W
W = -pext.dV = -50650 J
Q = -W = 50650J
deltaS = n.R.ln(Vfinal/Vini) = (1.0)(8.314)ln(2) = 5.763J/mol
A = U - TS => dA = dU -TdS => deltaA = 0 - T.deltaS = -298*5.763 = 1717J
G = H - TS => dG = dH - T.dS
dH = n.Cp.dT => dH = 0 for isothermal process
dG = -T.dS => deltaG = -T.deltaS = deltaA = 1717J
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madscientist

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Re:Isothermal expansion
« Reply #7 on: February 23, 2006, 03:48:20 PM »

Quote
so when u see w, it actually refers to the amount of work done by the system per unit mole of substance in the system. whereas W refer to the amount work done by the system.

what does a unit mol consist of?, and could you repeat what you did in the last post and include units so i can understand your method a bit better.

cheers,

madscientist :albert:
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Donaldson Tan

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Re:Isothermal expansion
« Reply #8 on: February 23, 2006, 04:02:14 PM »

so when u see w, it actually refers to the amount of work done by the system per unit mole of substance in the system. whereas W refer to the amount work done by the system.
The small capital denotes the value of the state function per unit mass. mass can be expressed in either in kg or moles. w can refer to amount of work done by the system per mole of substance or per kg of substance in the system.

With regards to your question on isothermal process, we will be working on a molar basis and not kg because you are provided the molar quantity (n=1.0mol) of the gas in interest.

It actually matters if the state function is express in terms of kg or moles, because the resultant equation will not be exactly the same.

let's consider molar basis for enthalpy of a perfect gas
h = u + pv = u + RT
unit of h = J/mol
unit of u = J/mol
unit of R = J/mol.K
unit of T = K

let's consider mass basis for enthalpy of a perfect gas
h = u + pv = u + RT/M where M is the molar mass of the gas
unit of h = J/kg
unit of u = J/kg
unit of R = J/mol.K
unit of T = K
unit of M = kg/mol
unit of RT/M = J/kg

from the above example, we can see that the equations for mass and molar basis are not almost entirely the same.
« Last Edit: February 23, 2006, 04:20:21 PM by geodome »
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madscientist

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Re:Isothermal expansion
« Reply #9 on: February 23, 2006, 04:21:39 PM »

Thanks geodome
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eutectic6002

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Re: Isothermal expansion
« Reply #10 on: August 31, 2006, 03:04:26 AM »

Hi everyone,
Would like to seek your help to clarify some doubts with regards to the questions posted by madscientist.

(a) isothermal, reversible expansion of an ideal gas
Question 1: For a reversible process, shouldn't delta G = 0?

(b) isothermal expansion of an ideal gas against a constant external pressure
Question 2: In this case, delta H = 0 but q is not equal to 0.
Since the external pressure is kept constant, shouldn't delta H = qp = q?

Question 3: G is a state function. Hence the value of delta G is path-independent. For (a) and (b), are the delta G values calculated the same?

Thank you.

Regards,
eutectic6002
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Yggdrasil

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Re: Isothermal expansion
« Reply #11 on: August 31, 2006, 05:48:49 AM »

(a) isothermal, reversible expansion of an ideal gas
Question 1: For a reversible process, shouldn't delta G = 0?

I think you may be confusing processes at equilibrium with reversible processes.  For a reaction at equilibrium, delta G = 0.  For a reversible process, delta G is not necessarily zero.

Quote
(b) isothermal expansion of an ideal gas against a constant external pressure
Question 2: In this case, delta H = 0 but q is not equal to 0.
Since the external pressure is kept constant, shouldn't delta H = qp = q?

I believe this applies only to reversible processes, since it doesn't make sense to relate a state function to a variable which is path dependent.

Quote
Question 3: G is a state function. Hence the value of delta G is path-independent. For (a) and (b), are the delta G values calculated the same?

They should be for the reason you gave.
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Yggdrasil

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Re: Isothermal expansion
« Reply #12 on: August 31, 2006, 06:06:13 AM »

After some though, I have a better answer for (b):

You have to make a distinction between Pext and Pint, that is the pressure of the surroundings and the pressure of the system.  delta H = qp only when the pressure of the system (Pint) stays constant.  For a reversible process, Pext = Pint, so you need not make the distinction here.  But, for irreversible proceses the two are not necessarily the same.  In the irreversible, isothermal expansion at constant external pressure, Pext stays constant, but Pint decreases throughout the expansion, until Pext = Pint.
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eutectic6002

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Re: Isothermal expansion
« Reply #13 on: August 31, 2006, 05:59:27 PM »

Hi Yggdrasil,
Thank you for the prompt reply. Really appreciate it. I’m still quite confused and welcome more inputs from you and others.

1.  Reversible processes, equilibrium and delta G
I tend to agree that delta G of reversible processes need not necessarily be zero after looking at many worked examples.

However I face difficulty explaining to students because in many textbooks and websites on Chemical Thermodynamics, it is often stated that a reversible process occurs in a series of infinitesimal steps, and at each step the system is at equilibrium with its surroundings. It is commonly stated that the entire reversible process is a succession of equilibrium states. In fact one reference even treats reversible processes as equilibrium processes (see below).

Reversible processes are also called equilibrium processes. The idea is that in a reversible process, such as a gas expansion, the system is always in balance (the external and internal pressures are only infinitesimally different) and so essentially in equilibrium. [Source: Page 21 of Energetics and Equilibria, Lecture notes from University of Cambridge]

Question 1
For the reversible isothermal expansion of an ideal gas, should I conclude that since the initial state and final state of the system are not at equilibrium hence delta G is not zero? This is despite the fact that the expansion is reversible with the system at equilibrium with its surroundings at each step in the expansion process.

Question 2
http://www.chem1.com/acad/webtext/thermeq/TE4.html
Problem example 2 in the above website deals with the reversible isothermal expansion of an ideal gas which is relevant to our discussion. In the given answers, delta G is not zero but delta S(universe) is zero. If delta S(universe) is zero, shouldn’t delta G = zero since delta G = –T delta S(universe)?

Question 3
Consider the isothermal expansion of an ideal gas (from V1 to V2) by the following three pathways:
Path 1: expansion into a vacuum
Path 2: expansion against constant external pressure
Path 3: reversible expansion

Since G is state function, the values of delta G obtained for the above process via different paths are the same implying that the maximum non-expansion work available via the different paths are the same. Is the preceding statement correct? 

2.  delta H = qp
I have checked many textbooks and websites and it seems that the constant pressure associated with the expression, delta H = qp, is the external pressure in which we carried out our experiments and not the pressure of the system. Hence I’m still puzzled by the fact that delta H = zero but q  is not 0 for the isothermal expansion of an ideal gas against a constant external pressure. In this case,  q is not qp?


Apologies for asking so much in such a messy way.

Thank you and regards,
eutectic6002

 
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Donaldson Tan

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Re: Isothermal expansion
« Reply #14 on: September 01, 2006, 04:08:40 AM »

Reversible processes are also called equilibrium processes. The idea is that in a reversible process, such as a gas expansion, the system is always in balance (the external and internal pressures are only infinitesimally different) and so essentially in equilibrium. [Source: Page 21 of Energetics and Equilibria, Lecture notes from University of Cambridge]

Question 1
For the reversible isothermal expansion of an ideal gas, should I conclude that since the initial state and final state of the system are not at equilibrium hence delta G is not zero? This is despite the fact that the expansion is reversible with the system at equilibrium with its surroundings at each step in the expansion process.

http://en.wikipedia.org/wiki/Quasistatic_equilibrium

Reversible processes are not necessary at equilibrium, but equilibrium is a good approximation to describe reversible processes. If these processes are really at equilibrium, then dG will actually be zero. We apply the same equations on reversible and equilibrium processes, but dG isn't always zero. Does this not indicate that reversible processes are not necessary at equilibrium?
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