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According to Newton's formula, the speed of the sound in air at STP (in $m/s$ ) is (Take the mass of 1 mole of air is $39 \times 10^{-6} \mathrm{~kg}$ )
Option: 1
$V=0.1406 \times 10^3 \mathrm{~m} / \mathrm{s}$

Option: 2
$V=0.2203 \times 10^3 \mathrm{~m} / \mathrm{s}$

Option: 3
$V=0.2406 \times 10^3 \mathrm{~m} / \mathrm{s}$

Option: 4
$V=0.2106 \times 10^3 \mathrm{~m} / \mathrm{s}$

Answers (1)

best_answer

1 mole of any gas occupies 22.4 litres at STP. Therefore, density of air at STP is

$p= \frac {Mass\, of\, one\, mole\, of\, air} {Volume\, of\, one\, male\, of\, air\, at\, STP}$

$=\frac{39 \times 10^{-3} \mathrm{~kg}}{22.4 \times 10^{-3}}=\frac{390}{224} \\$

$P=1.742 \mathrm{~kg} / \mathrm{m}^3$

At STP, $\rho=1 \mathrm{~at_m}=1.01 \times 10^5 \mathrm{~N} / \mathrm{m}^2$

According to Newton's formula, the speed of sound in air at STP is

$V=\sqrt{\frac{\rho}{p}} \\$

$V=\sqrt{\frac{1.01 \times 10^5}{1.742}} \\$

$V=\sqrt{\frac{101 \times 10^3 \times 10^5}{1742 \times 10^2}} \\$

$V=\frac{\sqrt{101}}{\sqrt{1742}} \times \sqrt{106} \\$

$V=\frac{10.04}{41.74} \times 10^3=\frac{1004}{4174} \times 10^3 \\$

$V=0.2406 \times 10^3 \mathrm{~m} / \mathrm{s}$

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Ritika Jonwal

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