# What is the change in entropy for an adiabatic expansion against an external pressure for an ideal gas?

##### 1 Answer

I assume you mean the entropy of the *system*. This won't be zero since the volume of the gas changes, assuming the expansion is reversible and that the gas is ideal.

If we are to examine

- Reversible work is
#w_(rev) = -int_(V_1)^(V_2) PdV# - Reversible heat flow is
#deltaq_(rev) = TdS#

In an *expansion*, the change in volume is *positive*, so that *heat flow between the system and surroundings is zero*, i.e.

By the **first law of thermodynamics**:

#DeltaU = cancel(q_(rev))^(0) + w_(rev) = -P int dV# ,

or in differential form:

#dU = cancel(deltaq_(rev))^(0) + deltaw_(rev) = -PdV# ,#" "" "bb((1))# where

#U# is the internal energy.

So all the work goes into changing the internal energy of the system, while *due to reversible heat flow* is zero.

However, the **change in entropy** can also be written as a function of temperature

#dS(T,V) = ((delS)/(delT))_VdT + ((delS)/(delV))_TdV# #" "" "bb((2))#

This treats the change in entropy as **two** steps:

- Change in entropy due to a change in
*temperature*at a constant*volume*. - Change in entropy due to a change in
*volume*at a constant*temperature*.

Note that the second term involves a change in entropy due to a change in volume, which is related to the work. With some manipulation using Maxwell's Relations, we end up with:

#dS(T,V) = 1/T((delU)/(delT))_VdT + ((delP)/(delT))_VdV# #" "" "bb((3))# where

#P# is pressure, and#T# is temperature.

For an ideal gas, we can evaluate using the ideal gas law that

#((delP)/(delT))_V = (del)/(delT)[(nRT)/V]_V = (nR)/V# ,

so:

#dS(T,V) = 1/T((delU)/(delT))_VdT + (nR)/VdV# #" "" "bb((4))#

Note that in an adiabatic process, we have that

#((delU)/(delT))_V = -Pcancel(((delV)/(delT))_V)^(0) = 0#

Treating the change in entropy in two steps, integrating

#DeltaS(T,V) = int_(T_1)^(T_2) 1/Tcancel(((delU)/(delT))_V)^(0)dT + int_(V_1)^(V_2) (nR)/VdV#

But the integral of zero is zero (no area under the curve!), so:

#color(blue)(DeltaS(T,V)) = 0 + color(blue)(nRln(V_2/V_1))#

Thus,