The calculations of ideal gas internal energy, enthalpy and entropy are explained in conjunction with the point of interest.
The proof of the isentropic relationship follows. Assume an ideal gas and a quasi-equilibrium or reversible process. Consider a differentially small movement of the piston in a piston-cylinder. Take a differential form of the Non Flow Energy Equation , noting that there is no heat transfer. $$ dU = dQ + dW = dW $$ Using IUPAC 1992 convention the work definition is ,
$$ dW = - p dV $$ For an Ideal Gas obtain the change in internal energy from the specific heat capacity at constant volume. Substitute into this the Ideal Gas Law $$ dU = m c_v dT = m c_v d (\frac{pV}{mR}) = \frac {c_v}{R} d(pV) = \frac{1}{\gamma-1} (pdV + Vdp) $$where \( \gamma = c_p/c_v \) is the ratio of heat capacities and c_p = c_v + R . Substitute expressions for \( dU \) and \(dW\) into the reduced form of the NFEE
$$ - p dV = \frac{1}{\gamma-1} (pdV + Vdp) $$ Separate variables (variables are p and V) to obtain $$ -\gamma p dV = V dp $$ $$ \gamma \frac{dV}{V} = - \frac{dp}{p} $$ Integrate from state 1 to current (c) pressure and volume $$ \gamma \int_1^c \frac{dV}{V} = - \int_1^c \frac{dp}{p} $$ $$ \gamma ln (\frac{V}{V_1}= -ln(\frac{p}{p_1}) $$ Exploitiong the Power Rule for Logarithms and rearranging we find, $$ p V^{\gamma} = p_1 V_1 ^{\gamma} = constant $$ This can be integrated to obtain the work of expansion (or compression). $$ W_{1,2} = - \int_1^2 p dV = -p_1 V_1 ^{\gamma} \int_1^2 \frac {1}{V^{\gamma}} dV $$ Under development, more to follow.