Adiabatic Change : An adiabatic change by definition, is one which does not allow any transfer of heat, i.e., Δq = 0 , it follows from the 1st law,

ΔU = − W

dU = − dW

Let only mechanical work of expansion or contraction is involved, dW = PdV. Moreover,

dU = CV dT

∴ CVdT = − PdV

For a system of 1 mole of an ideal gas, expanding adiabatically and reversibly from temp T1 to T2 and volume V1 to V2, we have

$\large C_v dT = -\frac{R T}{V} dV$

$\large \frac{dV}{V} = – \frac{C_v}{R} \frac{dT}{T}$

$\large \int_{V_1}^{V_2} \frac{dV}{V} = – \frac{C_v}{R} \int_{T_1}^{T_2} \frac{dT}{T}$

$\large ln\frac{V_2}{V_1} = – \frac{C_v}{R} ln\frac{T_2}{T_1}$

$\large ln\frac{V_2}{V_1} = \frac{C_v}{R} ln\frac{T_1}{T_2}$

$\large ln\frac{V_2}{V_1} = ln (\frac{T_1}{T_2})^{C_v/R}$

$\large V_1 T_1^{C_v/R} = V_2 T_2^{C_v/R} = K$

$\large C_p – C_v = R$

$\large ln\frac{V_2}{V_1} = \frac{C_v}{C_p -C_v} ln\frac{T_1}{T_2}$

$\large \frac{C_p – C_v}{C_v} ln\frac{V_2}{V_1} = ln\frac{T_1}{T_2}$

$\large (\gamma – 1) ln\frac{V_2}{V_1} = ln\frac{T_1}{T_2}$

$\large ln (\frac{V_2}{V_1})^{(\gamma – 1)} = ln\frac{T_1}{T_2}$

$\large (\frac{V_2}{V_1})^{(\gamma – 1)} = \frac{T_1}{T_2}$

$\large V_2^{\gamma – 1} T_2 = V_1^{\gamma – 1} T_1 = K$

$\large V^{\gamma – 1} T = K$

Substituting , $\large T = \frac{P V}{R}$

$\large V^{\gamma – 1} (\frac{P V}{R}) = K$

$\large P V^\gamma = Constant$

W = – Cv ΔT = −Cv ( T2 − T1) = Cv ( T1 − T2)

Where T1, T2 are initial and final temperatures.

For 1 mole of gas T = PV/R

Hence adiabatic work , $\large W = C_v (\frac{P_1 V_1}{R} – \frac{P_2 V_2}{R})$

$\large \frac{C_v}{R}(P_1 V_1 – P_2 V_2)$

$\large W = \frac{P_1 V_1 – P_2 V_2}{\gamma -1}$

Slope of PV curve in adiabatic & isothermal expansion:

For isothermal expansion of the gas, PV = K ⇒ P = V/K

The slope of the P V curve will be obtained from

$\large \frac{dP}{dV} = – \frac{K}{V^2} = – \frac{P}{V}$

For the adiabatic expansion of the gas PVγ = K

$\large P = \frac{K}{V^\gamma}$

$\large \frac{dP}{dV} = K (-\gamma)V^{-\gamma -1} = -\gamma \frac{P}{V}$

In the both the changes the slope is negative, since γ , is greater than 1 , the slope in the adiabatic P − V curve will be large than that in the isothermal one.