What is an adiabatic process ? Derive an expression for work done in an adiabatic process.

Explanation

The process during which there is no exchange of heat takes place between a system and its environment is known as an adiabatic process.
From first law of thermodynamics,
`DeltaQ=DeltaU+DeltaW, DeltaQ=0` for adiabatic process
`therefore DeltaU=-DeltaW` or `-DeltaU=DeltaW`
Mean work done by gas decreases the internal energy and work done on the gas increases the internal energy
For an ideal gas in adiabatic process,
`PV^gamma`=constant
Where `gamma` is the ratio of specific heats at constant pressure and at constant volume.
`therefore gamma=C_P/C_V`
The work done in an adiabatic change of an ideal gas from the state `[P_1,V_1,T_1]` to the state `[P_2,V_2,T_2]`
`DeltaW=PDeltaV`
The total work done W by total change,
`W=int_(V_1)^(V_2)PdV`...(1)
but `PV^gamma` = constant
`therefore P="constant"/V^gamma` ...(2)
`therefore` Putting the value of equ.(2) in equ.(1)
W=constant x `int_(V_1)^(V_2) (dV)/V^gamma`
=constant x `[V^(gamma+1)/(V_2-gamma+1)]_(V_1)^(V_2)`
`="constant"/(1-gamma)["constant"/V_2^(gamma-1)-"constant"/V_1^(gamma-1)]`
`=1/(1-gamma)[(P_2V_2^gamma)/V_2^(gamma-1) - (P_1V_1^gamma)/(gamma-1)]` [`because` From equation (2)]
`therefore W=1/(1-gamma)[P_2V_2^gammaxxV_2^(-gamma+1) - P_1V_1^gamma xx V_1^(-gamma+1)]`
`=1/(1-gamma)[P_2V_2-P_1V_1]`
but from equation of an ideal gas
`P_1V_1=muRT_1` and `P_2V_2=muRT_2`
`therefore W=1/(1-gamma)[muRT_2-muRT_1]`
`=1/(gamma-1)[muRT_1-muRT_2]`
`therefore W=(muR(T_1-T_2))/(gamma-1)` ...(3)
If work done by the gas in adiabatic process is W positive `(W gt 0)` then from equation (3),
`T_1-T_2 gt 0`
`therefore T_1 gt T_2` means temperature decreases
If work done on the gas in the process W is negative `(W lt 0)` then from equation (3)
`T_1 lt T_2 lt 0`
`therefore T_1 lt T_2` means the temperature of gas rises.