Cahllen Thermodynamics

(1) Let us say that you have an ideal gas in a sealed insulated container of volume Vo =Va + Vb,. The gas is held in the Va portion by a thin membrane while the Vb portion is evacuated. The temperature of the gas is measured to be Ti . Then, the membrane is ruptured (broken) and the gas fills the rest of the container. a. What is the final temperature? Explain why. b. Because the box is insulated there is no heat exchange with the outside. What is the change in entropy?


The box is sealed. The transformation is adiabatic (there is no heat exchange).

Considering the transformation cvasi-static we have

$\gamma = Cp/Cv$,   for mono atomic gas $\gamma = 5/3$
(for Hydrogen and Helium $\gamma =1.6$)

$T*[V^{(\gamma-1)}] = constant$

$Ti*Va^{(\gamma-1)} = Tf*(Va+Vb)^{(\gamma-1)}$

$Tf = Ti * [Va/(Va+Vb)]^{(\gamma-1)}$


If the adiabatic process is reversible then $\Delta(S) = 0$.

But if the expansion is sudden the the process is irreversible and $\Delta(S) > 0$.

$T*dS =dQv + P*dV$

$dS =(Cv*dT)/T + (P/T)*dV$             but we know that $PV = RT$
$dS = Cv* (dT/T) + R* (dV/V)$         now we integrate
$\Delta(S) = Cv*ln(T2/T1) + R*ln(V2/V1)$

$\Delta(S) = Cv*ln (V1/V2) + R ln(V2/V1) > 0$ for non reversible processes

In the case of a reversible and cvasistic process one has

$(V1/V2)^{(\gamma-1)} = T2/T1$     (see point a above)
and the variation in entropy is zero

$\Delta(S) = (\gamma-1)*Cv*ln(Va/V0) + R*ln(V0/Va) =$

$= [\gamma-1)*Cv – R]* ln(Va/V0) = [Cp-Cv-R]*ln(Va/V0) = (R-R)*ln(Va/V0) =0$