Adiabatic Process
It is the process where pressure, volume, and temperature of the system change but there is no exchange of heat between the system and the surrounding. The heat flow is prevented by the use of thermally insulating walls of quickly carrying out the process so that there is no enough time for attentional heat flow.
Examples of Adiabatic Process- Propagation of sound waves
- Sudden bursting of tubes of a bicycle tire
- The compression of stroke in an internal combustion engine
First Law applied to Adiabatic process
In the adiabatic process, there no change in heat energy in the system i.e. dQ = 0, from the first law
dQ = dU + dW
0 = dU + dW
dU = -dW
Equation of Adiabatic process
Consider n-mole of an ideal gas as a system enclosed in a cylindrical vessel with perfectly insulating walls and a movable frictionless piston. If it absorbs Q amount of heat, then it does dW amount work, which is related as
dQ = dU + dW …..... (i)
Since dU is only dependent on temperature,
dU = n Cv dT …..... (ii)
equation (i) becomes,
0 = n Cv dT + dW .....… (iii)
Since, dW = P dV where dV is the change in volume
0 = n Cv dT + P dV …..... (iv)
For n-mole of gas we have,
PV = nRT …..... (v)
Differentiating equation (v) and taking one mole, we get
dPV/dT = dRT/ dT
PdV + VdP = RdT
dT = (PdV + VdP)/R …..... (vi)
Substituting equation(vi) in equation (iv)
Cv (PdV + VdP) / R + P dV = 0
CvPdV + CvVdP + R PdV = 0
PdV (Cv + R) + CvVdP = 0
CpPdV + CvVdP = 0
Dividing both sides by CvPV, we get
(Cp/Cv). (dV/V) + (dP/P) = 0
Let the ratio Cp/Cv = γ, then
γ(dV/V) + (dP/P) = 0
Integrating both sides, we get
Similarly, for the relation of Temperature and volumeTVγ-1 = constant
For pressure and temperature
P1-γTγ = constant
Work done in Adiabatic Process
Consider the n-mole of a gas in a cylinder having perfectly non-conducting walls and fitted with a frictionless piston. Suppose the gas expands adiabatically from the initial volume V1 to the final volume V2. Then work done by the gas is
W = K [V1-γ/1-γ]v1 v2W = K/1-γ [V21-γ – V11-γ]
W = 1/1-γ [KV21-γ – KV11-γ]
Since, P1V1γ = P2V2γ = K, substituting the value and solving we get,
W = 1/1-γ (P2V2 – P1V1) ………... (i)
We know, PV= nRT, P1V1 = nRT1 and P2V2 = nRT2, putting these values in equation (i)
.: W = 1/1-γ (nRT2- nRT1)
= n. R/γ-1 (T1 – T2)
This equation given the work done for Adiabatic process
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