FL: Reversible and Irreversible Processes (take 2)

Adiabatic Reversible Expansion (Compression)

Consider the reversible expansion of an ideal gas where the system is now well-insulated (assume $c_v \ne F(T)$).

pistons schematic

$dU + dE_{k} + dE_{p} = \delta W + \delta Q$

$dU = \delta W + \delta Q$

$dU = \delta W$

pv rev ad expansion schematic

$dU = \delta W$

$\int c_v dT = - \int P_E dv$

Because the process is reversible $P_E \to P$, and that $d(RT) = d(Pv) = Pdv + vdP$, we can write

$d(RT) = d(Pv) = Pdv + vdP$

$dT = \frac{1}{R}Pdv + \frac{1}{R} vdP$

which we can plug back in to our original expression, to give

$\int \frac{c_v}{R} Pdv + \int \frac{c_v}{R}vdP = -\int Pdv$

$\int \frac{c_v}{R}vdP = -\int \left (1+\frac{c_v}{R} \right )Pdv$

$\int c_v vdP = -\int (R+c_v)Pdv = -\int c_P Pdv$

rearranging and integrating gives

$\int \frac{dP}{P} = -\int \frac{c_P}{c_v} \frac{dv}{v}$

$\ln\left (\frac{P_2}{P_1} \right ) = -k \ln \left (\frac{v_2}{v_1} \right ) = \ln \left (\frac{v_1}{v_2} \right )^k$

$\ln \left (P_1v_1^k \right ) = \ln \left (P_2v_2^k \right )$

$\left (Pv^k \right ) = constant$

$\left (PV^k \right ) = (different)constant$

Plugging this result back into the expression for $W$ (i.e., $W = - \int P_E dV = -\int P dV$

$W = -\int P dV = -\int (different)constant V^{-k} dV$

$W = \frac{(different)constant}{k-1}\left [\frac{1}{V_2^{k-1}}- \frac{1}{V_1^{k-1}} \right ]$

$W = \frac{1}{k-1}\left [P_2V_2-P_1V_1}\right ]$

$W = \frac{nR}{k-1}\left [T_2-T_1\right ]$


To obtain this result we assumed that $c_v$ (and $c_P$) were not functions of $T$.


How does the internal energy change differ for the compression?