What is the entropy change for 2 mols of a gas expanding from #"1 L"# to #"10 L"#?

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Answer 1 · 2017-12-12T19:27:39

You didn't mention it, so I have to assume you mean constant temperature... and I get #DeltaS = "38.3 J/K"# as a result.

And if so, then we calculate

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

where we consider the change in entropy due to the isothermal expansion of the gas.

The natural variables are #T# and #V#, so we consider the Helmholtz free energy Maxwell Relation:

#dA = -SdT - PdV#

For any state function, the cross-derivatives are equal, so:

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

Assuming an ideal gas is expanding, then we can use the ideal gas law...

#P = (nRT)/V#

Thus,

#((delP)/(delT))_V = (nR)/V((del(T))/(delT))_V = (nR)/V#,

and

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

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

#= color(green)(nRln(V_2/V_1))#.

Therefore, the change in entropy is:

#color(blue)(DeltaS) = "2 mols" cdot "8.314472 J/mol"cdot"K" cdot ln("10 L"/"1 L")#

#=# #color(blue)ul"38.3 J/K"#