**Problem: **

The mass of a hot air balloon and its cargo (not including the air inside) is
200 kg. The air outside is at 10 ^{o}C and 101 kPa. The
volume of the balloon is 400 m^{3}. To what temperature must the
air in the balloon be heated before the balloon will lift off. (Air
density at 10 ^{o}C is 1.25 kg/m^{3}.)

Solution:

- Concepts:

The buoyant force - Reasoning:

For the balloon to lift off, the buoyant force B must be greater than its weight. - Details of the calculation:

The buoyant force is equal to the weight of the displaced air at 10^{o}C = 283 K. B = (1.25 kg/m^{3})(400 m^{3})(9.8 m/s^{2}) = 4900 N. The weight of the balloon is 200 kg(9.8 m/s^{2}) + weight of hot air. The hot air therefore must weigh less than 4900 N - 1960 N = 2940 N. Its mass must be less than 2940 N/(9.8 m/s^{2}) = 300 kg. Its density must be less than r = 300 kg/(400 m^{3}) = 0.75 kg/m^{3}.

At constant pressure, the volume of a gas is proportional to the absolute temperature. (Law of Gay-Lussac) The pressures on the inside and outside of the inflated balloon are nearly equal. The pressure on the outside is the constant atmospheric pressure. The Law of Gay-Lussac therefore applies.

Since the volume of a gas at constant pressure is proportional to its temperature, its density ρ = m/V is proportional to 1/T.

We have ρ_{1}/ρ_{2 }= T_{2}/T_{1}. ρ_{1}T_{1}/ρ_{2 }= T_{2}. (1.25 kg/m^{3})(283 K)/(0.75 kg/m^{3}) = 472 K = T_{2}.

The air in the balloon must be heated to more than 472 K = 199^{o}C.

Two balloons have been filled up with air under atmospheric
pressure to volumes V_{1} and V_{2}, respectively. They
are now submerged under water. A thin string of length L, which is run
through a pulley at a fixed depth H, connects the balloons. (The radii of
the pulley and the balloons are much smaller than the length of the string.)
By setting the initial positions of the balloons, one can achieve a state of
equilibrium. Neither balloon is rising or going down. Determine the
difference in the depth of the balloons (in terms of H and L) under those
conditions. The mass of the balloon skins, of the string, and of the air
is negligible. The temperature of the water is constant and equal to the
temperature of the air.

Solution:

- Concepts:

The buoyant force, the ideal gas law, pressure at depth h, P_{h}= P_{top}+ ρgh

The ideal gas law: PV = NkT - Reasoning:

The buoyant force is equal to the weight of the displaced water. To experience the same buoyant force, the two balloons must have the same volume under water. - Details of the calculation:

At atmospheric pressure and at the same temperature we have V_{1}/V_{2}= N_{1}/N_{2}from the ideal gas law._{}To have the same volume under water at the same temperature we need P_{1}/P_{2}= N_{1}/N_{2}according to the ideal gas law. We therefore need

P_{1}/P_{2}= V_{1}/V_{2}.

(P_{top}+ ρgh_{1})/( P_{top}+ ρgh_{2}) = V_{1}/V_{2}._{}P_{top}+ ρgh_{1}= (V_{1}/V_{2})( P_{top}+ ρgh_{2}).

(1 – V_{1}/V_{2}) P_{top}+ ρgh_{1}= (V_{1}/V_{2})ρgh_{2}.

h_{2}= ((V_{2}/V_{1}) - 1)(P_{top}/ρg) + (V_{2}/V_{1})h_{1 }h_{2}– h_{1}= ((V_{2}/V_{1}) - 1)[(P_{top}/ρg) + h_{1})]

h

_{1}+ h_{2}= 2H - L

h_{2}– h_{1}= [(1 – (V_{1}/V_{2}))/ [(1 + (V_{1}/V_{2}))][2(P_{top}/ρg) + 2H – L]

Here P_{top}= 101 kPa and ρ = 1000kg/m^{3}.