A passenger plane flies at an altitude of 8300 m. To dispense with oxygen masks a constant pressure which corresponds to an altitude of 4150 m is maintained in the cabins with the aid of compressor. Take the mean temperature of outside air as equal to 0 degrees. The difference between pressure inside and outside the cabin is

(g = 10 m/s^2 , molar mass of air M = 28 * 10 ^ -3 kg , Gas constant R = 25/3 J/mol k , e = 2.72)

options

a) 2.4 atmos

b) 0.024 atmos

c) 0.24 atmos

d) 24 atmos

 

Pressure variation with heigh is given by

dP/dz = -dg (where dP/dz is differential and d is density and g is acc due to gravity in dg)

For air d= PM/RT

So dP/P = -(gM/RT)dz

gives P(z) =Po*exp(-gMz/(RT))

Where Po is pressure at surface = Patm

and P(z) is pressure at height z

 

Use this to find P(4150) and P(8300) and find difference. I have not done the exact calculation but I think it will come out to be 0.24 atm

(It can never be 2.4 and 24 atm as it must be less than 1 atm)