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

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}

Posted by

Ritika Jonwal

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