Why is the change in enthelpy 0 during an isothermal process?

[Xyius]

I am reviewing material for my thermodynamics class and one of the practice exams says that the change in enthalpy during an isothermal process is equal to zero. This doesn't make any sense to be considering..

\Delta H = \Delta U + PV

and since this is an isothermal process..

\Delta U = 0
Therefore..
\Delta H = PV \neq 0

?

 

[RedX]

I am reviewing material for my thermodynamics class and one of the practice exams says that the change in enthalpy during an isothermal process is equal to zero. This doesn't make any sense to be considering..

\Delta H = \Delta U + PV

and since this is an isothermal process..

\Delta U = 0
Therefore..
\Delta H = PV \neq 0

?

Well the real equation is:

\Delta H = \Delta U + (\Delta P )V+ P \Delta V

Now do some backwards investigation. If delta U is zero and delta H is zero, then it must be true that:

(\Delta P) V = - P \Delta V

or that:

\frac{\Delta V}{V}=-\frac{\Delta P}{P}

integrating:

V_f / V_i=P_i/P_f

doh, had a brain freeze. Basically: \Delta (PV)=0 is the condition for that to occur.

So basically your isothermal process conserves pressure times volume. If it's true for any process, I think one can conclude pressure times volume is a function of temperature alone. Dimensionally it would just have to be KT times a constant, unless a mass scale is involved, in which case it can be (mc²)ⁿ(kT)^1-n for any number n.

 

[DrDu]

Not even Delta U is zero in general in an isothermal process, with the exception of an ideal gas.

 

[nasu]

If
H=U+PV
you should know without any further calculations that for an isothermal process in an ideal gas, U is constant and PV is constant (Boyle's law). So H will constant or deltaH=0.

 

[Xyius]

Ohh! Okay that makes sense, I forgot that I was dealing with an ideal gas. (Even though I didn't explicitly say so in the post.) I got bogged down because I was thinking of the general case. Thanks everyone!

 

[nasu]

I don't think this is a result valid in general. Even for real gases the enthalpy depends on temperature. So the question is likely to refer to ideal gas.

 

[Studiot]

Going back to the original question, remember the first law anyone

dU = 0 maybe, but
dU = q + w, neither of which may be zero.

So for an isothermal reversible expansion

\begin{array}{l}<br /> dU = dq - PdV = {\left( {\frac{{\partial U}}{{\partial U}}} \right)_V} + {\left( {\frac{{\partial U}}{{\partial V}}} \right)_T}dV = 0 \\ <br /> dq = - dw = Pdv \\ <br /> SinceP = \frac{{nRT}}{V} \\ <br /> \int {dq = - \int {dw} } = \int {nRT\frac{{dV}}{V}} = nRT\ln \frac{{{V_2}}}{{{V_1}}} = nRT\ln \frac{{{P_1}}}{{{P_2}}} \\ <br /> \end{array}

Further
Is condensation not a valid isothermal process for an ideal gas?

 

[nasu]

Going back to the original question, remember the first law anyone

dU = 0 maybe, but
dU = q + w, neither of which may be zero.

So for an isothermal reversible expansion

\begin{array}{l}<br /> dU = dq - PdV = {\left( {\frac{{\partial U}}{{\partial U}}} \right)_V} + {\left( {\frac{{\partial U}}{{\partial V}}} \right)_T}dV = 0 \\ <br /> dq = - dw = Pdv \\ <br /> SinceP = \frac{{nRT}}{V} \\ <br /> \int {dq = - \int {dw} } = \int {nRT\frac{{dV}}{V}} = nRT\ln \frac{{{V_2}}}{{{V_1}}} = nRT\ln \frac{{{P_1}}}{{{P_2}}} \\ <br /> \end{array}

Further
Is condensation not a valid isothermal process for an ideal gas?

What do you mean by condensation? The gas-liquid transition?

As for the demonstration regarding the work in isothermal process, I don't understand how is related to the original question. Can you explain?

 

[Studiot]

1) Yes I mean the gas to liquid phase transition, liberating the latent heat of boiling.

2) My point is that the equation in the OP contains an explicit work term, but the heat term is lost in (combined with) other quantities.

 

[nasu]

1) Yes I mean the gas to liquid phase transition, liberating the latent heat of boiling.

OK, then this does not apply to ideal gas. There is no condensation predicted by the ideal gas model. You need to use more realistic models (like Van der Waals model) to see any condensation.

2) My point is that the equation in the OP contains an explicit work term, but the heat term is lost in (combined with) other quantities.

You mean this one: H=U+PV ?
One of the reason it is used is that for processes at constant pressure (like many chemical processes) the change in enthalpy is equal to the heat transferred.
Not that it matters for the OP question which is very straightforward.

 

[Studiot]

You mean this one: H=U+PV ?

No I read post#1 properly.
That was the equation you introduced.

I would rather let my major point that even if \DeltaU is zero it's components as given by the First Law will probably not be, come to the fore.

 

[nasu]

No I read post#1 properly.

I would rather let my major point that even if \DeltaU is zero it's components as given by the First Law will probably not be, come to the fore.

They will not be zero, of course. It they were both zero (heat, work) then nothing much can happen. Does it look like I assumed that they are both zero? Or maybe someone else here?

 

[Redbelly98]

Are we still wondering about the question in the OP? It was resolved in Post #'s 4 and 5. The OP meant to refer to an ideal gas, and Post #4 showed why it is true.

Is condensation not a valid isothermal process for an ideal gas?

No. The condensed phase is not a gas, let alone an ideal one.

2) My point is that the equation in the OP contains an explicit work term, but the heat term is lost in (combined with) other quantities.

I don't see the explicit work term. Do you mean "PV"? That is not work.

If your point is that Q and W aren't necessarily zero in an isothermal process, I don't think anyone claimed that they are. But that seems irrelevant to the OP. Maybe I am not seeing your point.