Answer:
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.
