Enthelpy change for an ideal gas

AI Thread Summary
The discussion revolves around the equation ΔH = ΔU + (Δn)RT, with confusion regarding the application of temperature (T) in the context of ideal gases. It is clarified that if temperature is constant, ΔU can be considered zero, leading to ΔH = (Δn)RT. The participants also discuss how Δn represents the change in the number of moles of gaseous products minus reactants, which affects the enthalpy change. Additionally, the relationship between internal energy (U) and temperature (T) is emphasized, noting that U is proportional to both T and the number of moles (n). The conversation ultimately resolves the initial confusion about the equation and its implications for ideal gases.
babita
Messages
61
Reaction score
0

Homework Statement



I'm not able to understand the following equation
ΔH = ΔU + (Δn)RT
firstly if T is taken to be constant (as the book says), ΔU = 0
if T is not constant then which T i am supposed to put in? initial or final?

Homework Equations


please help. Thank you.


The Attempt at a Solution

 
Physics news on Phys.org
babita said:

Homework Statement



I'm not able to understand the following equation
ΔH = ΔU + (Δn)RT
firstly if T is taken to be constant (as the book says), ΔU = 0
if T is not constant then which T i am supposed to put in? initial or final?

Homework Equations


please help. Thank you.

The Attempt at a Solution


Hi babita! :smile:

Looks to me that your equation is not right.
I think it should be:
ΔH = ΔU + Δ(nRT)

Assuming n is constant, this is the same as:
ΔH = ΔU + nRΔT

Does that answer your question?
To give a more extensive explanation:

H is defined as H=U+PV.
With the ideal gas law PV=nRT, it follows that H=U+nRT.
For a change in H we get:
ΔH=Δ(U+nRT)=ΔU+Δ(nRT)
 
hi:smile:
yeah that would have made sense but its written "at constant temperature" every where :'(
 
Okay, so apparently the amount of matter does not stay constant and you have a Δn.

If the temperature is constant then the initial temperature is the same as the final temperature.

Your equation becomes:
ΔH=ΔU+Δ(nRT)=ΔU+(Δn)RT.

And as you surmised, with T constant, you have ΔU=0, so you get:
ΔH=(Δn)RT
 
I like Serena said:
Okay, so apparently the amount of matter does not stay constant and you have a Δn.
amount of matter may or may not change...Δn means no of moles of gaseous products minus no of moles of gaseous reactants

THAT is my confusion...at constant T , ΔU makes no sense
 
Actually, in retrospect ΔU does make sense if the number of moles changes.
My bad.

For an ideal gas you have: U=n Cv T
With constant T, the change in U is:
ΔU=(Δn) Cv T
 
Cv is heat capacity at constant volume...i don't think volume is constant here...in my book the equation have been derived assuming constant T & P.
 
Also Internal energy of an ideal gas is directly proportional to T
 
Cv is indeed the heat capacity at constant volume.

However, it turns out that the formula U=n Cv T holds for an ideal gas, even if the volume is not constant.

That's not really relevant here though.
As you said internal energy U is directly proportional to T.
U is also directly proportional to the number of moles n.
 
  • #10
I like Serena said:
Cv is indeed the heat capacity at constant volume.

However, it turns out that the formula U=n Cv T holds for an ideal gas, even if the volume is not constant.

.

yeah sry ...that was silly
 
  • #11
and yes U is proportional to n, equation makes sense at constant T ...missed that point... thanks :)
 
  • #12
You're welcome. :)
 
Back
Top