Question

A weather balloon rises through the atmosphere, its volume expanding from 3.4^3 to 11^3 as the temperature drops from 23 ∘C to -12 ∘C

If the initial gas pressure inside the balloon is 1.0 atmatm, what is the final pressure?

Note: the answer is not 0.032/0.32/0.30/0.03.

Answer 1

Using Ideal gaw law:

PV = nRT

Since number of moles will remain constant inside balloon, So

PV/T = Constant

P1*V1/T1 = P2*V2/T2

P2 = final pressure = P1*V1*T2/(T1*V2)

Using given values:

P1 = 1.0 atm = 1.01325*10^5 Pa

V1 = Initial Volume = 3.4 m^3

V2 = final volume = 11 m^3

T1 = Initial temperature = 23 C = 273 + 23 = 296 K

T2 = final temperature = -12 C = 273 - 12 = 261 K

So,

P2 = 1.01325*10^5*3.4*261/(296*11)

P2 = 27615.4 Pa

P2 = (27615.4 Pa)*(1 atm/(1.01325*10^5))

P2 = 27615.4/(1.01325*10^5) atm

P2 = 0.272543 atm

In two significant figures

P2 = final pressure = 0.27 atm

Let me know if you've any query.