Question

Large helium-filled balloons are used to lift scientific equipment to high altitudes.

(a)What is the pressure inside such a balloon (in atm) if it starts out at sea level with a temperature of 24.8°C and rises to an altitude where its volume is sixteen times the original volume and its temperature is −33.7°C? (Enter your answer to at least 3 decimal places)

atm

(b)What is the gauge pressure? (Enter your answer in atm and to at least 3 decimal places. Assume atmospheric pressure is constant.)

atm

Answer 1

Part A.

Using Ideal Gas law:

PV = nRT

PV/T = nR = Constant

So,

P1V1/T1 = P2V2/T2

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

Given that

P1 = 1.0 atm = 1.01325*10^5 N/m^2

V1 = V

V2 = 16*V

T1 = 24.8 C = 24.8 + 273 = 297.8 K

T2 = -33.7 C = 273 - 33.7 = 239.3 K

So,

P2 = 1.01325*10^5*V*239.3/(16*V*297.8)

P2 = 1.01325*10^5*239.3/(16*297.8)

P2 = 5088.8 N/m^2

P2 = (5088.8 N/m^2)*(1 atm/(1.01325*10^5 N/m^2))

P2 = 0.05022 atm = 5.022*10^-2 atm

Part B.

Gauge Pressure is given by:

Gauge Pressure = Actual Pressure - P_atm

Gauge Pressure = 0.05022 atm - 1.00 atm

Gauge Pressure = -0.94978 atm = -0.950 atm

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