Suppose we wanted to heat the air in a flexable room-size balloon from 60F to 65
ID: 801831 • Letter: S
Question
Suppose we wanted to heat the air in a flexable room-size balloon from 60F to 65F. The balloon is tethered at a ground level in Albuquerque, its initial volume is 100 cubic meter, and its initial pressure is equal to the ambient pressure of 83000Pa. Because the balloon is flexable, it can expand, but the pressure will remain the same in order to maintain hydrostatic balance with the surrounding atmosphere. The effects of the balloon surface tension may be ignored.
a). How much heat has to be supplied to heat the balloon by 5F(2.8K)?
b). How much will the balloon volume change (hint: use the ideal gas law).
3. Suppose that an air parcel at the surface has a pressure of 100000Pa and a temperature of 302.9K. It is forced to rise along a dry adiabat and reaches saturation at an altitude of 1 km.
a). What is the temperature of the air parcel at 1 km?
b.) What is the pressure in the air parcel?
Hint: for all questions with potential temperature, please keep mind: "Potential temperature is a conserved quantity for a air parcel that moves around in the atmosphere under adiabatic conditions."
Explanation / Answer
a.Difference of temperature=65-60=5F=2.8 K
Specific capacity of air=1.006KJ/kgC
Volume of air =100m3=3531.47cufeet
Since mass of air=Density *volume
Density of air=0.037kg/cu.feet
Mass of air=0.037*3531.47=130.6kg
Heat required=specific heat capacity*mass of air*temperature difference
=1.0006*130.6*2.8
=365.89KJ
b. According to Charles law
V1/T1/V2/T2
V1= initial volume=100m3
V2=??
T1=60F=288.7K
T2=65 F=291.4K
Now,
100/V2=288/291.4
100*291.4/288=V2
29140/288=V2
101.1m3
3.a. Surface air pressure of gas=100000pa
temperature=302.9K
adiabatic=dT/dz=0.098*(degree celsius/m)
=0.098degree celsius/m*(-dz)
=(0.098degree celius)*(-1km)*(1000m/1km)
=-9.8degree celsius=282.95 K
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