Air is pumped into the tubes of a cycle rickshaw at a pressure of $2$ atm. The volume of each tube at this pressure is $0.002 m^{3}$ . One of the tubes gets punctured and the volume of the tube reduces to $0.0005 m^{3}$ . How many moles of air have leaked out? Assume that the temperature remains constant at $300 K$ and that the air behaves as an ideal gas.

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Solution

The pressure of the air is $P_{1} = 2 a t m = 2 \times 10^{5} p a$
The volume of the tube is $V_{1} = 0.002 m^{3}$
The constant temperature of the tube is $T_{1} = 300 K$

Using the ideal gas relation:
$P_{1} V_{1} = n_{1} R T_{1}$

$\Rightarrow n = \frac{P_{1} V_{1}}{R T_{1}}$

$= \frac{2 \times 10^{5} \times 0.002}{8.3 \times 300} = \frac{4}{8.3 \times 3} = 0.1606$

Now, after the tube gets punctured,
$P_{2} = 1 a t m = 10^{5} p a$
$V_{2} = 0.005 m^{3}$ ,
$T_{2} = 300 K$

Using the ideal gas relation again:
$P_{2} V_{2} = n_{2} R T_{2}$

$\Rightarrow n_{2} = \frac{P_{2} V_{2}}{R T_{2}}$

$= \frac{10^{5} \times 0.0005}{8.3 \times 300}$

$= \frac{5}{3 \times 8.3} \times \frac{1}{10} = 0.02$

$Δ n =$ moles leaked out $= 0.16 - 0.02 = 0.14$

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