# How do you calculate the change in entropy for an isothermal process?

##### 1 Answer

It depends on which way we change the entropy (pressure or volume?).

Pick one, and you would use one of these:

#DeltaS = nRln(V_2/V_1)#

#DeltaS = -nRln(P_2/P_1)#

Actually, if you notice that at constant

#ln(V_2/V_1) = ln((P_1cancel(V_1)"/"P_2)/(cancel(V_1)))#

#= ln[(P_2/P_1)^(-1)] = -ln(P_2/P_1)# ,

which means our two expressions are equivalent for any monatomic ideal gas.

**DISCLAIMER:** *This contains Calculus. I assume that if you read further, you understand partial derivatives, cross-derivatives, and the integral of*

**Entropy** is typically considered a function of temperature and either volume or pressure. When we ** hold temperature constant** (an isothermal process), and change one of the other parameters:

(1)#DeltaS = int_(P_1)^(P_2) ((delS)/(delP))_TdP#

(2)#DeltaS = int_(V_1)^(V_2) ((delS)/(delV))_TdV#

If you notice, we don't have an expression for the way entropy changes due to pressure or volume in relation to gases. It would be nice to have a way to get these relationships from known formulas, like the ideal gas law.

It turns out that these can be derived from the **Maxwell Relations** that contain

(3)#dG = -SdT + VdP# ,

where

(4)#dA = -SdT - PdV# ,

where

**ENTROPY AS A FUNCTION OF TEMPERATURE AND PRESSURE**

The first context in which we calculate **(3)**; here is how you should examine

#" "" "" "uarr "------------------------------------" darr#

#dG = stackrel("Numerator")overbrace(-S)" "stackrel("Hold constant")overbrace(dT) + V" "stackrel("Denominator")overbrace(dP)#

#" "" "" "" "" "" "" "uarr "---------------------" darr#

Do something similar for

#((delS)/(delP))_T = -((delV)/(delT))_P#

Then, for a monatomic ideal gas, we can use the **ideal gas law** (

#V = (nRT)/P#

#=> color(green)(-((delV)/(delT))_P) = -(del)/(delT)[(nRT)/P]_P#

#= -1/P(d)/(dT)[nRT] = color(green)(-(nR)/P)#

That means our first way to get the change in entropy is (plugging back into **(1)**):

#color(blue)(DeltaS) = int_(P_1)^(P_2) ((delS)/(delP))_TdP#

#= -int_(P_1)^(P_2) ((delV)/(delT))_PdP#

#= -int_(P_1)^(P_2) (nR)/PdP = -nRint_(P_1)^(P_2) 1/PdP#

#= color(blue)(-nRln(P_2/P_1))#

**ENTROPY AS A FUNCTION OF TEMPERATURE AND VOLUME**

The other way to express **(2)** and **(4)**:

#" "" "" "uarr "------------------------------------" darr#

#dA = stackrel("Numerator")overbrace(-S)" "stackrel("Hold constant")overbrace(dT) - P" "stackrel("Denominator")overbrace(dV)#

#" "" "" "" "" "" "" "uarr "---------------------" darr#

Do something similar for

#((delS)/(delV))_T = ((delP)/(delT))_V#

On the ideal gas law, this means:

#P = (nRT)/V#

#=> color(green)(((delP)/(delT))_V) = (del)/(delT)[(nRT)/V]_V#

#= 1/V(d)/(dT)[nRT] = color(green)((nR)/V)#

Substitute this result back into **(2)** to get:

#color(blue)(DeltaS) = int_(V_1)^(V_2) ((delS)/(delV))_TdV#

#= int_(V_1)^(V_2) ((delP)/(delT))_VdV#

#= int_(V_1)^(V_2) (nR)/VdV = nRint_(V_1)^(V_2) 1/VdV#

#= color(blue)(nRln(V_2/V_1))#

*Overall, if you know the initial and final pressure, or the initial and final volume, you can figure out the change in entropy of a monatomic ideal gas at a constant temperature.*