A container in the shape of a cube 10.0 cm on each edge contains air (with equivalent molar mass 28.9 g/mol) at atmospheric pressure and temperature 300 K. Find (a) the mass of the gas, (b) the gravitational force exerted on it, and (c) the force it exerts on each face of the cube. (d) Why does such a small sample exert such a great force? (6%)

a) m = 1.174 grams

b) F_g = 0.01151 N

c) F_c = 1013 N

Explanation:

Given:

- The length of a cube L = 10.0 cm

- The molar mass of air M = 28.9 g/mol

- Pressure of air P = 101.3 KPa

- Temperature of air T = 300 K

- Universal Gas constant R = 8.314 J/kgK

Find:

(a) the mass of the gas

(b) the gravitational force exerted on it

(c) the force it exerts on each face of the cube

(d) Why does such a small sample exert such a great force? (6%)

Solution:

- Compute the volume of the cube:

V = L^3 = 0.1^3 = 0.001 m^3

- Use Ideal gas law equation and compute number of moles of air n:

P*V = n*R*T

n = P*V / R*T

n = 101.3*10^3 * 0.001 / 8.314*300

n = 0.04061 moles

- Compute the mass of the gas:

m = n*M

m = 0.04061*28.9

m = 1.174 grams

- The gravitational force exerted on the mass of gas is due to its weight:

F_g = m*g

F_g = 1.174*9.81*10^-3

F_g = 0.01151 N

- The force exerted on each face of cube is due its surface area:

F_c = P*A

F_c = (101.3*10^3) * (0.1) ^2

F_c = 1013 N

- The molecules of a gas have high kinetic energy; hence, high momentum. When they collide with the walls they transfer momentum per unit time as force. Higher the velocity of the particles higher the momentum higher the force exerted.