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
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