enthalpy-internal energy and work done (doubts )

1) starting with this one-work done
dW=PextdV but in all the cases we take Pext=nRT/V in solving the problems,we integrate from there but Pext is not equal to nRT/V??? then why are we taking equal???

2) internal energy-
Does 'U' change even when the temperature is constant???and why???

3)Enthalpy-
E=U+PV and is heat supplied at constant pressure,So for An Adiabatic change at constant Pressure it should be zero but an IIT 2004 question on this topic doesnt consider one

And the last one-gimme some clarity about how we can make an constant pressure adiabatic change???

I Need the answers ASAP.plz

1 Answers

21
Swaraj Dalmia ·

1)We only take Pext=nRT/V in reversible process i.e. Pext≈Pint=nRT/V

2)For cont. temp U remains const. only for a closed or isolated process,where there is no change of mass.
For open systems,more no. of molecules at same temp. may be added,this increases the total int. energy.

3)Adiabatic process is a process where heat exchanged is zero.i.e. there is interaction between work done and int. energy.
A simple example of constant pressure adiabatic change is a quick[adiabatic] irreversible expansion against const. external pressure.Here work is done at the cost of int. energy.

The 3rd one is interesting:
Now from 1st law of thermo-d.
Q=U+W [heat supplied=inc. in int. energy+work done]
Work done can be of 2 types:
useful:like electric[nFe]
non-useful:PV work.

Therefore we can rewrite it as:
Q=U+PVwork+Wuseful ....(i)

Note that the statement:
Q=U+PVwork
is wrong,as it does not include all work done.

For Adiabatic process Q=0.
Now,equating E=U+PV & (i)
E=-Wuseful
Hence the result.

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