# Thermodynamics Concepts

1. Mar 23, 2011

### theown1

Hi, I have a couple of questions for my general knowledge because I have a test coming up and I'm confused on some stuff

1) For an isothermal process is the $$\Delta$$S=0 because $$\Delta$$T=0?
since it equals zero does that mean that the process is irreversible?

2)What makes a process reversible? is that when the $$\Delta$$S>0?

3)In an adiabatic process, pV$$\gamma$$=const. means that it equals a constant value not it is constant because if the temperature changes then the pressure and volume or one of them has to change right? But I dont understand how that is constant when Q=0, but then when $$\Delta$$T=0, just PV is constant, wouldn't what I had earlier be constant at $$\Delta$$T=0 too?

4) In an adiabatic process when I use piVi$$\gamma$$=pfVf$$\gamma$$, means that I can just find the final pressure or the volume after the process occurs?

5)Similarly what does this mean? THVi$$\gamma$$-1=TLVf$$\gamma$$-1

and if I was only given the heat transfer QH and QL could I substitute that in for TH and TL?

6)In an Isochoric process, the volume is constant, so to find Q or $$\Delta$$U, I can use either =nCv$$\Delta$$T or =3/2nR$$\Delta$$T?

7)In an Isobaric process, constant pressure, does $$\Delta$$U=Q? Is this correct Q=mc$$\Delta$$T or $$\Delta$$U=nCv$$\Delta$$T? or do I use something with Cp like $$\Delta$$U=nCp$$\Delta$$T?

2. Mar 23, 2011

### Andrew Mason

No. Isothermal processes inevitably involve change in entropy. If the internal energy does not change then volume must change (otherwise, there is no change at all). The reversible path between the initial and final state must involve work and, therefore, heat flow: dQ = dW; dS = dQ/T.

A reversible process can have a positive or negative change in entropy of the system depending on whether the heat flow is into or out of the system. However, the entropy change of the system + surroundings will always be 0. A reversible process is one that occurs while the system and surroundings are in thermal and mechanical equilibrium (ie. out of equilibriium by an infinitessimal amount).

The temperature, pressure and volume all change in an adiabatic process. $PV^\gamma = K$ tells you how P changes relative to V. From that you can work out T: PV=nRT.

T is not constant in an adiabatic change. Volume changes so work is done. Q = 0 so $\Delta U = -W$

And T. Once you have found P and V you can find T.

Just substitute P = nRT/V in $PV^\gamma = K$

No.

Only if it is a monatomic gas.

For a constant pressure process dQ = nCpdT = dU + dW = nCvdT + PdV = nCvdT + nRdT = n(Cv+R)dT

AM

Last edited: Mar 23, 2011
3. Mar 24, 2011

### theown1

thanks that cleared up a lot!