A football has a diameter of 22 cm. It is inflated to 1 atm pressure above normal atmospheric pressure on a sunny summer day when the temperature was 25C. Take normal atmospheric pressure to be 101.3 kPa. What mass of air is contained in the ball?

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Explanation:

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Dec 10, 2017

See below. The answer is about $13.20$ g.

Explanation:

We are going to use the formula(the ideal gas law) $P V = n R T$.

$P$: pressure (Pa)
$V$: volume (${m}^{3}$)
$n$: amount of substance(mol)
$R$: gas constant($8.314$ $J m o {l}^{-} 1 {K}^{-} 1$)
$T$: absolute temparature(K)

Then, what is the value of the variants?
[1] $P = 2$ atm $= 2.026 \cdot {10}^{5}$ Pa
(I interpreted "inflated to 1 atm pressure above normal atmospheric pressure" to be 2 atm. If this is wrong, let me know.)
[2] $V = \frac{4}{3} \pi \cdot {0.11}^{3} = 5.575 \cdot {10}^{-} 3$ ${m}^{3}$.
[3] $T = 25$$= 298.15$ K.

Plug in these values to the formula.
$n = \frac{P V}{R T} = \frac{2.026 \cdot {10}^{5} \cdot 5.575 \cdot {10}^{-} 3}{8.314 \cdot 298.15}$
$= 0.4557$ mol.

The avarage molar mass of the air is $28.966$ g/mol (From Japanese wikipedia: https://ja.wikipedia.org/wiki/%E7%A9%BA%E6%B0%97), and $0.4557$ mole of air is $28.966 \cdot 0.4557 = 13.199$ grams.

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