Question 4
(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]
Reference: Past Exam Paper – June 2016 Paper 42 Q2
Solution:
(a)
(i)
The Avogadro constant NA 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
Thank you. Kindly solve Q2 from 42/m/j/17 too.
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