**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 C

_{v}dT …..... (ii)equation (i) becomes,

0 = n C

_{v}dT + dW .....… (iii)Since, dW = P dV where dV is the change in volume

0 = n C

_{v}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)

C

_{v}(PdV + VdP) / R + P dV = 0C

_{v}PdV + C_{v}VdP + R PdV = 0PdV (Cv + R) + CvVdP = 0

C

_{p}PdV + CvVdP = 0Dividing both sides by C

_{v}PV, we get(C

_{p}/C_{v}). (dV/V) + (dP/P) = 0Let the ratio C

_{p}/C_{v }= γ, thenγ(dV/V) + (dP/P) = 0

Integrating both sides, we get

Similarly, for the relation of Temperature and volume

TV

^{γ-1}= constantFor pressure and temperature

P

^{1-γ}T^{γ}= 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 V

_{1}to the final volume V_{2}. Then work done by the gas isW = K [V

^{1-γ}/1-γ]_{v1 }^{v2}W = K/1-γ [V

_{2}^{1-γ}– V_{1}^{1-γ}]W = 1/1-γ [KV

_{2}^{1-γ}– KV_{1}^{1-γ}]Since, P

_{1}V_{1}^{γ}= P_{2}V_{2}^{γ}= K, substituting the value and solving we get,W = 1/1-γ (P

_{2}V_{2}– P_{1}V_{1}) ………... (i)We know, PV= nRT, P

_{1}V_{1}= nRT_{1}and P_{2}V_{2}= nRT_{2}, putting these values in equation (i).: W = 1/1-γ (nRT

_{2}- nRT_{1}) = n. R/γ-1 (T

_{1}– T_{2})This equation given the work done for Adiabatic process

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