At a pressure of 1.05 × 105 Pa and a temperature of 27 °C, 1.00 mol of helium gas has a volume of 0.0240 m3.

Question 3

(a) Describe the motion of molecules in a gas, according to the kinetic theory of gases. [2]

(b) Describe what is observed when viewing Brownian motion that provides evidence for your answer in (a). [2]

(c) At a pressure of 1.05 × 10⁵ Pa and a temperature of 27 °C, 1.00 mol of helium gas has a volume of 0.0240 m³.

The mass of 1.00 mol of helium gas, assumed to be an ideal gas, is 4.00 g.

(i) Calculate the root-mean-square (r.m.s.) speed of an atom of helium gas for a temperature of 27 °C. [3]

(ii) Using your answer in (i), calculate the r.m.s. speed of the atoms at 177 °C. [3]

[Total: 10]

Reference: Past Exam Paper – June 2017 Paper 41 & 43 Q4

Solution:

(a) The molecules move in random motion with a distribution of speeds and in different directions.

(b) Small specks of light are observed to move in random directions.

(c)

(i)

pV = ⅓ Nm〈c²〉

{N is the total number of molecules; N = 1 (for 1 atom)

and m is the mass of 1 molecule

So, Nm is the total mass of the gas (in kg)}

1.05×10⁵ × 0.0240 = ⅓ × 4.00×10^-3 × 〈c²〉

〈c²〉 = 1.89 × 10⁶           

OR

{Kinetic energy = (3 / 2) kT }

½ m〈c²〉 = (3 / 2) kT

{m is the mass of 1 molecule / atom

1 mole - - - > 4.00 g

6.02 × 10²³ molecules - - > 4.00 g = 4.00 × 10^-3 kg

1 molecule - - > (4.00 × 10^-3 / 6.02 × 10²³)

Temperature should be in kelvin}

0.5 × (4.00 × 10^-3 / 6.02 × 10²³) × 〈c²〉 = 1.5 × 1.38 × 10^-23 × 300

〈c²〉 = 1.89 × 10⁶

OR

{pV = nRT                   and pV = 1/3 Nm〈c²〉

So,}

nRT = ⅓ Nm〈c²〉

{n : number of moles

N: total number of molecules

m: mass of 1 molecule

Nm: total mass of gas}

1.00 × 8.31 × 300 = ⅓ × 4.00 × 10^-3 × 〈c²〉

〈c²〉 = 1.89 × 10⁶           

{c_rms = √〈c²〉}

cr.m.s. = 1.37 × 10³ m s^-1

(ii)

{ nRT = ⅓ Nm〈c²〉

So, 〈c2〉 ∝ T }

〈c2〉 ∝ T          

{The mean-square speed is proportional to the absolute temperature.

T is the thermodynamic temperature (that in, in kelvin)

27°C = 300 K              and 177°C = 177 + 273 = 450 K

From above,

〈c2〉 ∝ T

〈c2〉 at 300 K - - > 1.89×10⁶

〈c2〉 at 450 K - - > 1.89 × 10⁶ × (450 / 300)}

〈c2〉 at 177 °C = 1.89 × 10⁶ × (450 / 300)

cr.m.s. at 177 °C = 1.68 × 10³ m s^-1