1.

Derive the equation of adiabatic changes

Answer»

Consider 1 g mole of n ideal gas enclosed in a cylinder fitted with a perfectly friction less piston. Let P, V, T be initial pressure, volume and temperature.

Suppose a small amount of heat dQ spent in two ways:

Increasing the temperature, of gas by small range dT at constant volume = cvdT

Expansion of gas by small volume dV portion of heat spend = PdV

dQ= CvdT + PdV

In adiabatic change

dQ = 0 CvdT + PdV = 0   …(1) 

According to standard gas equation 

PV = RT 

Differentiate both sides, 

PdV + VdP = RdT 

dT = \(\frac{PdV+VdP}{R}\) 

Putting eq. (1), 

\(\frac{C_v(PdV+VdP)}{R}\) + PdV = 0 

(Cv + R)PdV + CvVdP = 0  …(2) 

But Cp − Cv = R 

From (2) dividing both sides by CvPV \(\frac{C_PdV}{C_vPV}+\frac{C_vVdV}{C_vPV}=0\) 

\(\frac{\gamma dV}{v}+\frac{dP}{P}\)

γ =\(\frac{C_p}{C_v}\) 

Integrating both sides. 

Γ∫ \(\frac{dV}{V}\) + ∫ \(\frac{dP}{P}\) = C 

γloge V + logeP = C  

logeVγ + logeP = C  

LogkPVγ = C  

PVγ = K This is the required relation


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