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Saturday, July 20, 2019

Gas molecules leak from the container in (b) at a constant rate of 1.5 × 1019 s-1. The temperature remains at 17 °C.


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

2 comments:

  1. Thank you. Kindly solve Q2 from 42/m/j/17 too.

    ReplyDelete
    Replies
    1. go to
      http://physics-ref.blogspot.com/2019/07/the-pressure-p-and-volume-v-of-ideal.html

      Delete

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