Answer

If the Low pressure in Hurricane Andrew was 922 mb and density

remained constant (which it wouldn't) at 1.225 kg m^-3

^{what is the temperature in K and deg C?}

^{P =922 mb - convert this to Pa >>> 92200
Pa}

^{I don't have Greek Letters so I am going to use
rho to indicate density rather than the Greek letter}

^{P = Pressure (Pa)}

^{rho = density (kg} m^{-3 )}

R = Gas constant ( 287 J kg^-1K)

^{T = Temperature ( K)}

^{P = (rho) R T}

^Therefore

^{T = P / ((rho) R)}

T= 92200 Pa / ((1.225kg m^-3) x (287J kg^-1K)) = 262.2 K

Then convert Temperature K to Temperature deg C

T= 262.2 K - 273.2 = -10.9 ^oC

^{What will happen to the volume
of a balloon when it leaves the surface? Demonstrate this using the ideal
gas law.}

^{The Volume will increase.}

^{Remember that density = mass/volume}

^{P = (rho) RT = (mass/volume) R T}

^{m = mass}

^{V = volume}

^{The mass will remain constant as the balloon rises.}

^{mass=constant}

^{R= constant - by definition}

^{If we assume T remains constant ( which we know
it won't - what will it do? - Hint see above question)}

^{T = constant}

^{That leaves Pressure - We know that Pressure decreases
with height}

^{P = (m/V) RT}

^{P/ (RT) = m/V}

^{Invert both sides}

^{(RT)/ P = V /m}

^{Move m across to the LHS (left hand side)}

^{m ((RT)/P) = V}

From above - we have m, R,T are constants so we are left with

P and V

From the equation we can see that P is inversely related to V

Try this with some numbers e.g. P = 1000 mb (i.e. close to sea level)

and P = 900 mb - i.e. higher into the atmosphere.

(You could use a Temperature of 20 deg C and m =1.22 kg)

At P = 1000 mb V= 1.0259 m³

At P =900 mb V = 1.1399 m³

^{Hints if you are not getting these numerical answers}

^{a) Units - remember to convert Pressure into Pa}

^{- remember to convert Temperature into K}

^{b) Don't forget the Gas constant}