Question

Derive the relation between temperature and pressure for a reversible adiabatic expansion of an ideal gas

Answer 1

When you first posted this question I was very busy, but now I've found the time to look at it again and can give you an answer:

As I already mentioned in one of my comments your equation is slightly incorrect. It should actually be

T2=[cV+Rp2p1cp]T1T2=[cV+Rp2p1cp]T1

Now, I will come to the derivation: You should always start out by identifying what you already know about the system. You know the expansion will be performed adiabatically, so there can be not heat exchange with the environment, i.e.

δQ=0 .δQ=0 .

This means that from the first law of thermodynamics you get:

dU=δQ+δW=δWdU=δQ+δW=δW

where UU is the internal energy and WW is the work. Furthermore, you know that the expansion process is irreversible. If the only work performed is the pressure-volume work, where the system has to expand against a constant external pressure, which is p2p2 in your equation, then this work is given by:

δW=−p2dVwhere:p2=const.≠f(V)δW=−p2dVwhere:p2=const.≠f(V)

So, together with the first law of thermodynamics this yields:

dU=−p2dV .(1)(1)dU=−p2dV .

Now, you need to assume that the expanding gas is an ideal gas. For ideal gases you know that

dUdU=∂U∂T=CVdT+∂U∂V=Π=0dV=CVdT(2)dU=∂U∂T⏟=CVdT+∂U∂V⏟=Π=0dV(2)dU=CVdT

where TT is the temperature, CVCV is the heat capacity at constant volume, and ΠΠ is the internal pressure, which is equal to zero for an ideal gas. Other relations that hold for ideal gases are the ideal gas law

pV=nRTpV=nRT

where nn is the amount of the gas and RR is the universal gas constant, and

cp−cV=R(3)(3)cp−cV=R

where cp=Cpncp=Cpn and cV=CVncV=CVn are the molar heat capacities at constant pressure and constant volume, respectively.

Now, you just have to combine equations (1)(1) and (2)(2):

−p2dV=CVdT ,−p2dV=CVdT ,

integrate the result (under the assumption that CVCV is temperature independent)

−p2∫V2V1dV−p2(V2−V1)=CV∫T2T1dT=CV(T2−T1) ,−p2∫V1V2dV=CV∫T1T2dT−p2(V2−V1)=CV(T2−T1) ,

use the ideal gas law V=nRTpV=nRTp to substitute pp for VV

−nRp2(T2p2−T1p1)−nR(T2−p2p1T1)=CV(T2−T1)=CV(T2−T1) ,−nRp2(T2p2−T1p1)=CV(T2−T1)−nR(T2−p2p1T1)=CV(T2−T1) ,

introduce the molar heat capacity via CV=ncVCV=ncV

−R(T2−p2p1T1)=cV(T2−T1) ,−R(T2−p2p1T1)=cV(T2−T1) ,

and separate T1T1 and T2T2

T1(cV+p2p1R)=T2(cV+R) .T1(cV+p2p1R)=T2(cV+R) .

Finally, you can use equation (3)(3) to get the desired result:

T1(cV+p2p1R)T1(cV+p2p1R)⇒ T2=T2(cV+R=cp)=T2cp=[cV+Rp2p1cp]T1 .

[Thank You]