State Function Conditions in Isothermal, adiabatic

[cns]

In an Isothermic and reversible condition why is DH=0? (D=delta) I can understand that DU=0 but DH? but why would DH not change?

DH should also =0 if it was an adiabtic and irreversible condition? since DH=dq/T -> as long as there is no heat transfer DH=0?

Since it's reversible would DG=0? or is it because it's isothermal? how would DG change for adiabatic, isochor, isobaric condtions?

Thanks so much

 

[Andrew Mason]

In an Isothermic and reversible condition why is DH=0? (D=delta) I can understand that DU=0 but DH? but why would DH not change?

Do you mean an isothermic and reversible process? $\Delta H$ is usually used for change in enthalpy. Do you mean enthalpy or entropy?

If the process is isothermal, there is a non-zero change in entropy. This is because dU = 0 so dQ = dW = PdV which must be non-zero in a reversible process. so dS = dQ/T <> 0.

DH should also =0 if it was an adiabtic and irreversible condition? since DH=dq/T -> as long as there is no heat transfer DH=0?

In an adiabatic reversible process, there is no heat flow so dS = dQ/T = 0. If the adiabatic process is not reversible, there is a change in entropy. This is because entropy is a state function that is determined by taking the integral of dS over the reversible path between two states. (If the path actually taken is an irreversible adiabatic path, the reversible path from the beginning state to that end state will not be adiabatic - heat will flow so ds<>0).

Since it's reversible would DG=0? or is it because it's isothermal? how would DG change for adiabatic, isochor, isobaric condtions?

What is dG?

AM

 

[cns]

sorry, I should have defined them

dH= Enthalpy
dG=free energy

My main question is why is DH=0 when you have an isothermal and reversible process against an external pressure of 1 atm?

thanks

 

[Andrew Mason]

In an Isothermic and reversible condition why is DH=0? (D=delta) I can understand that DU=0 but DH? but why would DH not change?

The heat flow is not zero in a reversible isothermal process. Why do you say it is?

DH should also =0 if it was an adiabtic and irreversible condition? since DH=dq/T -> as long as there is no heat transfer DH=0?

As I explained in my earlier post and in https://www.physicsforums.com/showpost.php?p=2959987&postcount=4", $\Delta H$ is zero in an adiabatic process, whether it is reversible or irreversible. But we calculate change in entropy $\Delta S$ always along the reversible path between two states. If the actual adiabatic path between two states is not reversible, the reversible path between those two states will involve heat flow, so dQ/T is going to be non-zero.

AM

 

[Andrew Mason]

Since it's reversible would DG=0? or is it because it's isothermal? how would DG change for adiabatic, isochor, isobaric condtions?

G = H - TS = (U + PV) - TS, so

$$dG = dU + PdV + VdP - TdS - SdT$$

If the process isothermal and isobaric, then dP = dT = dU = 0, so

dG = PdV - TdSSince dS = dQ/T,

For a reversible isothermal path between the two states, it follows that

$$\Delta Q = T\Delta S$$

For an isothermal and isobaric path between two states dU = 0 so, from the first law, $\Delta Q = P\Delta V$

So if the path is reversible, isobaric and isothermal:

$$P\Delta V - T\Delta S = 0$$

Since $\Delta G = P\Delta V - T\Delta S$, $\Delta G = 0$ for a reversible isothermal and isobaric process.

AM

 

[cns]

Thanks Andrew,

This is the proof in my book for isothermal conditions for why DU and DH = 0

DH=n*Cp*ln(v2/v1)*(T2-T1)

since no change in temperature DH = 0

Is this correct?

because I thought the DH = q so DH shouldn't = 0.

thanks again

 

[Andrew Mason]

Perhaps you could show us the entire proof. Thanks

AM