Adiabatic Process (Reversible)

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 $

Adiabatic work:

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.

Also Read :

→Thermodynamic Process
→ Thermodynamic Process
→ Differential form of the First Law
→ Workdone in Thermodynamics
→ Heat Capacity
→ Thermochemistry

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