Thermodynamics

This is a case of adiabatic expansion. For adiabatic expansion, following equations must be satisfied :

PVγ = Constant where P, V are the pressure and volume and γ = Cp/Cv

Or P1V1γ = P2V2γ

Or (P1/ P2) = (V2/V1)γ .......(1)

And assuming air to be an ideal gas : PV = nRT

Or P1V1 = nRT1 and P2V2 = nRT2

Or (V2/V1) = (P1/P2)(T2/T1) ......(2)

Substituting from (2) in (1) : (P1/ P2) = (P1/P2)γ(T2/T1)γ

Or (P1/ P2)1-γ = (T2/T1)γ

Or (P1/ P2)(1-γ)/γ = (T2/T1) ........(3)

For air Cp = 28.86 J/mole.K

Also we know that Cp - Cv = R or 28.86 - Cv = 8.31 or Cv = 20.55 J/mole.K

Hence, γ = Cp/Cv = 28.86/20.55 = 1.4

Substituting P1 = 0.802 atm, P2 = 0.602 atm, T1 = 288 K and γ = 1.4 in (3) we get :

(0.802/0.602)(1-1.4)/1.4 = T2/288

Or 1.332-0.286 = T2/288

Or 1.3320.286 = 288/T2

Taking log on both the sides : 0.286 log101.332 = log10288 - log10T2

Or log10T2 = 2.424 or T2 = 265 K (or - 8OC)

As the temperature falls to -8OC (below the freezing point of water), it will snow.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!