Question

A container in the shape of a cube 11.6 cm on each edge contains air (with equivalent molar mass 28.9 g/mol) at atmospheric pressure and temperature 291 K. (a) Find the mass of the gas. kg (b) Find the gravitational force exerted on it. mN (c) Find the force it exerts on each face of the cube. kN (d) Why does such a small sample exert such a great force

Answer 1

a. 0.00189 kg

b. 18.552 mN

c. 1.363 kN

d. Since the molecular density is high, the force exerted by the sample is thus high.

Step-by-step explanation:

(a) Find the mass of the gas. kg

Using PV = mRT/M where P = pressure on gas = atmospheric pressure = 1.013 × 10⁵ Pa, V = volume of gas = L³ where L = length of cube = 11.6 cm = 0.116 cm,m = mas of gas, R = molar gas constant = 8.314 J/mol-K, T = temperature of gas = 291 K and M = molar mass of gas = 28.9 g/mol

So, m = PVM/RT = PL³M/RT

Substituting the values of the variables into the equation, we have

m = PL³M/RT

= 1.013 × 10⁵ Pa × (0.116)³ × 28.9 g/mol/ 8.314 J/mol-K × 291 K

= 0.0457 × 10⁵ Pa g/mol/2419.374J/mol

= 1.89 × 10⁻⁵ × 10⁵ g

= 1.89 g

= 1.89 × 10⁻³kg

= 0.00189 kg

(b) Find the gravitational force exerted on it. mN

The gravitational force, F exerted on it is its weight W

So, F = W = mg where m = mass of gas = 1.89 × 10⁻³ kg and g = acceleration due to gravity = 9.8 m/s²

F = mg

= 1.89 × 10⁻³ kg × 9.8 m/s²

= 18.522 × 10⁻³ kgm/s²

= 18.552 × 10⁻³ N

= 18.552 mN

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

Since pressure, P = F/A where F = force exerted on each face and A = area of each face = L² where L = length of side of cube = 11.6 cm = 0.116 m

So, F = PA since P = atmospheric pressure = 1.013 × 10⁵ Pa,

F = PL²

= 1.013 × 10⁵ Pa (0.116 m)²

= 0.01363 × 10⁵ N

= 1.363 × 10³ N

= 1.363 kN

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

To answer this question, we need to find the density of the gas in the cube.

So density of gas,ρ = m/V where m = mass of gas = 1.89 g and V = volume of gas = L³ and L = length of side of cube = 11.6 cm

ρ = m/V = m/L³ = 1.89 g/(11.6 cm)³ = 1.89 g/1560.896 cm³ = 0.00121 g/cm³

We now find the number of moles of gas in a cm³ by dividing its density by its molar mass.

So n = ρ/M = 0.00121 g/cm³ ÷ 28.9 g/mol = 23687.67 mol/cm³

Since there are 6.022 × 10²³/mol, we find the number of molecules in a cm³ which is n × 6.022 × 10²³/mol = 23687.67 mol/cm³ × 6.022 × 10²³/mol

= 143731.1 × 10²³ molecules/cm³

= 1.437311 × 10²⁸ molecules/cm³

≅ 1.44 × 10²⁸ molecules/cm³

Since the molecular density is high, the force exerted by the sample is thus high.