A room has dimensions of 5.0 m X 4.5 m X 4.0 m.

(a) What weight of air occupies the room when the air pressure is 1.01 X 10

^{5} Pa and the density is 1.15 kgm

^{-3}?

(b) A typical scuba tank with volume 1.1 X 10

^{-2} m

^{2} can be pressured up to 2.0 X 10

^{7} Pa. How many scuba tanks will it take to store this amount of air?

(b) What is the magnitude of the atmosphere’s downward force on your phone, which we take to have an area of 4.0 X 10

^{-3} m

^{2}?

Density of air = 1.15 kgm^{-3}

Volume of room = 5.0 m X 4.5 m X 4.0 m

Volume of room = 90 m^{3}

Volume of scuba tank = 1.1 X 10^{-2} m^{2}

Weight = mg

Mass = ρV

Weight = ρVg

(a) Weight of air in room = 1.15 kgm^{-3} X 90 m^{3} X 9.8 m^{3}s^{-2}

Weight of air in room = 1014N

(b) P_{1}V_{1} = P_{2}V_{2} if T is constant.

Volume of air in room if compressed to scuba tank = [(1.01 X 10^{6} Pa) X (90 m^{3}) / (2.0 X 10^{7} Pa)]

Volume of air in room if compressed to scuba tank = 4.545 m^{3}

No of scuba tanks required = [(1.01 X 10^{5} Pa) X (90 m^{3}) / (2.0 X 10^{7} Pa)] / [1.1 X 10^{-2} m^{2}]

No of scuba tanks required = 42 tanks.

(c) Force = Pressure X Area

Force on phone face =
(1.01 X 10^{5} Pa) X (4.0 X 10^{-3} m^{2})

Force on phone face = 404N