(a) Explain what is meant by the Avogadro constant. (b) Argon-40 (4018Ar) may

Q#2 (Past Exam Paper – June 2014 Paper 41 & 43 Q2)

(a) Explain what is meant by the Avogadro constant. [2]

(b) Argon-40 (⁴⁰18Ar) may be assumed to be an ideal gas.

A mass of 3.2 g of argon-40 has a volume of 210 cm³ at a temperature of 37 °C.

Determine, for this mass of argon-40 gas,

(i) the amount, in mol, [1]

(ii) the pressure, [2]

(iii) the root-mean-square (r.m.s.) speed of an argon atom. [3]

Solution:

(a) The Avogadro constant is the number of atoms in 12g of carbon-12.

(b)

(i)

{The nucleon number (mass number) represents the mass of 1 mole of the element.

40 g --> 1 mol of Ar

3.2 g --> (3.2/40) mol}

Amount = 3.2 / 40 = 0.080 mol

(ii)

{Ideal gas equation : pV = nRT

All quantities should be in SI units. The SI unit of temperature is K.

T = 273 + 37 = 310 K

1 cm³ = 10^-6 m³}

p × 210×10^-6 = 0.080 × 8.31 × 310

Pressure, p = 9.8×10⁵ Pa

(iii)

EITHER

pV = (1/3) Nm<c²>

{where N is the number of atoms and m is the mass of 1 atom. So, Nm is the total mass.

1 mol contains 6.02×10²³ atoms.

0.080 mol contains 0.080 × 6.02×10²³ atoms.}

Number of atoms, N = 0.080 × (6.02×10²³) [= 4.82×10²²]

{The Ar atom contains 40 nucleon.

Mass of 1 nucleon (u) = 1.66×10^-27 kg

Mass of 1 atom of Ar (mass of 40 nucleons) = 40 × 1.66×10^-27 kg}

and  mass of 1 Ar atom, m = 40 × (1.66×10^-27) [= 6.64×10^-26]

{ pV = (1/3) Nm<c²> }

9.8×10⁵ × 210×10^-6 = (1/3) × 4.82×10²² × 6.64×10^-26 × <c²>

<c²> = 1.93×10⁵

r.m.s. speed, c_RMS = √<c²> = 440 m s^-1

OR

{As given in the question, mass = 3.2 g = 3.2×10^-3 kg}

(Total mass of gas) Nm = 3.2×10^-3    

{ pV = (1/3) Nm<c²> }

9.8×10⁵ × 210×10^-6 = (1/3) × 3.2×10^-3 × <c²>           

<c²> = 1.93×10⁵

r.m.s. speed, c_RMS = √<c²> = 440 m s^-1

OR

{Kinetic energy = 3/2 kT}

½ m<c²> = (3/2) kT    

{Mass m is the mass of Ar (= mass of 40 nucleons = 40u)}

½ × (40 × 1.66×10^-27) × <c²> = (3/2) × 1.38×10^-23 × 310    

  

<c²> = 1.93×10⁵ 

r.m.s. speed, c_RMS = √<c²> = 440 m s^-1