# Relation between Pressure and Volume during an Adiabatic Process

## Relation between Pressure and Volume during an Adiabatic Process

When a thermodynamic system undergoes a change in such a way that no exchange of heat takes place between system and surroundings, the process is known as adiabatic process.

In this process Pressure (P), Volume (V) and Temperature (T) changes but ΔQ = 0.

Let, n moles of a gas be made to expand from an initial state (P₁V₁) to a final state (P₂V₂) adiabatically.

From dU + dW = 0 at any state,

dU = nCVdT and dW = PdV

∴ nCVdT + PdV = 0

Let us eliminate ndT, by making use of the equation of state from PV = nRT.

Differentiating and remembering that P, V and T all of them changes.

PdV + VdP = nRdT

ndT = 1/R (PdV + VdP)

The above equation reduces to $$\frac{{{C}_{V}}(PdV+VdP)}{R}+PdV=0$$.

(CV + R) PdV + CVVdP = 0

CVPdV + CVVdP = 0 [From Mayer’s relation].

Dividing throughout by CVVP:

$$\frac{{{C}_{P}}}{{{C}_{V}}}\frac{dV}{V}+\frac{dP}{P}=0$$.

Substituting $$\frac{{{C}_{P}}}{{{C}_{V}}}=\gamma$$ a constant known as the ratio of specific heats of a gas, we get:

$$\gamma \frac{dV}{V}+\frac{dP}{P}=0$$.

$$\gamma \int\limits_{{{V}_{1}}}^{{{V}_{2}}}{\frac{dV}{V}}=-\int\limits_{{{P}_{1}}}^{{{P}_{2}}}{\frac{dP}{P}}$$.

$$\gamma \ln \left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)=-\ln \left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)=\ln \left( \frac{{{P}_{1}}}{{{P}_{2}}} \right)$$ (Or) $${{\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)}^{\gamma }}=\frac{{{P}_{1}}}{{{P}_{2}}}$$.

P₁V₁γ = P₂V₂γ

i.e. PVγ = k (Constant).