FOLLOW US ON TWITTER
SHARE THIS PAGE ON FACEBOOK, TWITTER, WHATSAPP ... USING THE BUTTONS ON THE LEFT


YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Friday, December 28, 2018

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


Question 2
(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]





Reference: Past Exam Paper – June 2014 Paper 41 & 43 Q2





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, cRMS = <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, cRMS = <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, cRMS = <c2> = 440 m s-1

No comments:

Post a Comment

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | A mass of 3.2 g of argon-40 has a volume of 210 cm3 at a temperature of 37 °C.