(a) State what is meant by (i) the Avogadro constant NA, (ii) the mole.

Q#4 (Past Exam Paper – June 2016 Paper 42 Q2)

(a) State what is meant by

(i) the Avogadro constant NA, [1]

(ii) the mole. [2]

(b) A container has a volume of 1.8 × 10⁴ cm³.

The ideal gas in the container has a pressure of 2.0 × 10⁷ Pa at a temperature of 17 °C.

Show that the amount of gas in the cylinder is 150 mol. [1]

(c) Gas molecules leak from the container in (b) at a constant rate of 1.5 × 10¹⁹ s^-1.

The temperature remains at 17 °C.

In a time t, the amount of gas in the container is found to be reduced by 5.0%.

Calculate

(i) the pressure of the gas after the time t, [2]

(ii) the time t. [3]

[Total: 9]

Solution:

(a)

(i) The Avogadro constant N_A is the number of atoms/nuclei in 12 g of carbon-12.

(ii) The mole is the amount of substance which contains the same number of atoms as there are in 12 g of carbon-12.

(b)

pV = nRT

{1 cm³ = 10^-6 m³

17 °C = 17 + 273 = 290 K}

2.0×10⁷ × 1.8×10⁴×10^-6 = n × 8.31 × 290

so n = 149 mol or 150 mol

(c)

(i)

{As the gas molecules leak from the container, the number of molecules (or number of moles) decreases with time.

pV = nRT

The temperature T remains constant.

The gas occupies the volume of the container, so V is also constant.

R is already a constant.

So, p ∝ n

The pressure is proportional to the number of moles.}

V and T constant and so pressure reduced by 5.0%

{The amount of gas is reduced by 5.0 %. So, only (100 – 5 =) 95 % of the original amount remains.

We have should that the pressure is proportional to the number of moles. So, if the amount is now 95 % of the initial amount, the pressure is also 95 % of the original one.}

Pressure = 0.95 × 2.0 × 10⁷

pressure = 1.9 × 10⁷ Pa

(ii)

{The amount of molecules is reduced by 5.0 %.

The loss corresponds to}

loss = 5 / 100 × 150 mol = 7.5 mol

or

{in terms of number of molecules,

Loss in number of molecules = 7.5 × 6.02×10²³}

ΔN = 4.52 × 10²⁴

{The gas leaks at a constant rate of 1.5×10¹⁹ s^-1. That is, in 1 second, 1.5×10¹⁹ molecules leak.

Now, we find the time required for (7.5 × 6.02×10²³) molecules to leak.}

t = (7.5 × 6.02×10²³) / 1.5×10¹⁹

or

t = 4.52 × 10²⁴ / 1.5 × 10¹⁹

t = 3.0 × 10⁵ s

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