Hello sir/ma'am
My doubt is that for an isothermal reversible process why should ∆H = 0? Shouldn't it have some other value because there is a change in volume?( ∆H=∆U+P∆V, ∆U=0 )

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Solution

Well basically the form of the enthalpy is that

dH=dE+d(pV)(not only pdV.)

dH=dE+pdV+Vdp

For the isothermal process dE=0

And for using the ideal gas equation pV=mRT____(1)

Differentiating (1) with v we can write pdV=constant because T is constant

And again differentiating equation (1) with p we can write Vdp=0

So that last two terms are zero ,too.

So that for ideal gas we can say that the enthalpy is zero

So in one sentence 'the pV terms becoms the function of constant T'

Yes. Isothermal process have enthalpy change.

But, enthalpy is not only a function of internal energy or temprature.

Change in enthalpy = change in internal energy + change in PV.

In case of isothermal process:

Change in internal energy = 0. And if change in PV is zero enthalpy change is also zero.

It appears that PV change is zero for isothermal process, that is true only for ideal gases. For real gases, it is not so and thus for real gases there is a some enthalpy change when it undergoes an isothermal process.

So, there isn't any contradiction.

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Similar questions

At constant pressure (∆P=0)

Than, Enthalpy

∆H=∆U+∆(PV)

∆H=∆U+P∆V+V∆P; But as (∆P=0)

∆H=∆U+P∆V

But if the Volume is constant (∆V=0)

Than the equation becomes

∆H=∆U+V∆P

What does this Equation implies and what is it's application?

Determine the value of ∆H and ∆U for the reversible isothermal evaporation of 90.0g of water at 100°C. Assume that water vapour behave as am ideal gas and heat of eveporation of water is 540cal.

MY PROBLEM

Only tell me How ∆H is calculated here, because if I know∆H then I can use the equation

∆H = ∆U + nRT

In my book it is given that

∆H = 90.0 x 540

= 48600 cal

But I want to know that why they have multipled the mass to heat of eveporation to find out ∆H

Q. For a Chemical Process at constant temperature and pressure,

Δ H

is Equal to

Q. For gaseous reactions, if

Δ H

is the change in enthalpy and

Δ U

that in internal energy, then

Δ H

is always greater than

Δ U

.
If true 1 else 0.

Q. For which reaction,

Δ H > Δ U

? (

Δ H =

change in enthalpy,

Δ U =

change in internal energy )

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Hello sir/ma'am My doubt is that for an isothermal reversible process why should ∆H = 0? Shouldn't it have some other value because there is a change in volume?( ∆H=∆U+P∆V, ∆U=0 )

Hello sir/ma'am
My doubt is that for an isothermal reversible process why should ∆H = 0? Shouldn't it have some other value because there is a change in volume?( ∆H=∆U+P∆V, ∆U=0 )