[Mel22]

A gas is allowed to expand instantaneously against a pressure which is half it's original value, the temperature of the thermostat it is contained in being remained at 600K.  Given that the heat supplied is 2.49kJ , determine the entropy change of the thermostat and show whether or not this instantaneous expansion is reversible.

Ok, so on can calculate

delta S (Thermostat)= -2.49kJ/600K = -4.15J/K

if dS = dS(sys) + dS(th)=0

then the process is reversible


that is straight forward, however:

delta S(System) = q(system)/T = w'/T      where w'= q ( as T is constant)

w'= p(external)*dV

however: how do I know dV and p?

Does anyone see what has gone wrong here? THANK YOU SO MUCH!

PS: Also see attached file, if this is not clear

[madscientist]

Hi Mel,

I downloaded that pdf and had a decent look and to me it looks like you need to have a look at something called the clausius inequality
( if you havnt allready) this law states:

dS is greater than or equal to dq/T

the clausius ineqaulity applies to irreversible, spontaneous processes,

for eg.) Delta S > q/T ,  

this states that the entropy change will greater than q/T for an irreversible spontaneous process

the clausius eqaulity applies to reversible processes,

for eg.) Delta S = q/T


Now for the question your looking at, ( (d) I pressume),
for the external pressure which is stated as being one half the original pressure, this is actually equal to the final pressure of the system,  therefore the pressure ratio is equal to 0.5.

I.E)   Pfinal / Pinitial = 0.5

      *(It doesnt matter what the actual quantitys are)
     
now entropy change can be calculated by using the following equation  

I.E) dS= - (n R ln (Pfinal / Pinitial))
         
          = -(1mol)(8.314 JK.mol)ln(0.5)

          = -(-5.763 JK)

          = 5.763 JK

now looking at the clausius

dS = q /T

     = 2.49kJ/600K

     = 4.15J/K

so looking at the actual calculated enthalpy change value:

dS = 5.763 JK > q/T

the dS value is greater than q/t, and therefore the process is irreversible and spontaneous

I could be wrong with this so if anyone can confirm that this is right (or more probably wrong) it would help both Mel and me!

cheers,

madscientist :albert:

[Mel22]

Thank you very much madscientist!I think you are probably right. At least I came up with the same approach after having had another thought about it.

As far as I understand it, the Caussian Inequality is essentially just dSuniv >= dS(Sys) + d(Ssurr) written in another form.

Can anyone confirm that what madscientist and me did is right? That would be great! :-)

Cheers:1eye:

[Mel22]

Can anyone please tell me whether that is correct? That would be really helpful! Thanks so much!  

[Mel22]

Also, in the formula dS= - (n R ln (Pfinal / Pinitial))
doesn't p refer to the internal instead of the external pressure???

Maybe one could do it like this:

dS = p(ext)(dV)/T

where p(ext) is inversely proprtional to V (pV = nRT), so therefore if the pressure halves the Volume doubles....substituting this in gives

dS = 0.5 p * 2 dV /T = pdV/T

 where pdV =q= 2.49 kJ /T   (derived in an eralier part of the question)

which gives dS (system) = -dS (thermostat)

therefore the process is reversible....however, I don't think it is reversible (just a feeling)...can anyone help please?

[Yggdrasil]

Also, in the formula dS= - (n R ln (Pfinal / Pinitial))
doesn't p refer to the internal instead of the external pressure???

Maybe one could do it like this:

dS = p(ext)(dV)/T

where p(ext) is inversely proprtional to V (pV = nRT), so therefore if the pressure halves the Volume doubles....substituting this in gives

dS = 0.5 p * 2 dV /T = pdV/T

 where pdV =q= 2.49 kJ /T   (derived in an eralier part of the question)

which gives dS (system) = -dS (thermostat)

therefore the process is reversible....however, I don't think it is reversible (just a feeling)...can anyone help please?

Recall that dS = dq(rev)/dT.  Even though the process under consideration is irreversible, you must calculate dS using a reversible path (so P(ext)=P).  Since entropy is a state function, the value of DeltaS is path independent.  So as long as your reversible path has the same initial and final states as your irreversible path, the two processes will have the same DeltaS.

[Mel22]

thanks for this...so am I right in the assumpotion that madscientist's approach is correct?

I'll be off and read another book on Thermodynamics now... I really want to understand this and I'm sure it can't be THAT hard!