The temperature at which the velocity of sound in oxygen will be the same as that of nitrogen at $15^{\circ} C$ is

112^{\circ} C

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72^{\circ} C

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56^{\circ} C

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17^{\circ} C

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Solution

The correct option is C $56^{\circ} C$
$V_{s} = \sqrt{\frac{γ R T}{M}}$
$M_{O_{2}} = 32$ , $M_{N_{2}} = 24$
$V_{O_{2}} at T = V_{N_{2}} at 288 K$ ,
$T = \frac{M_{O_{2}}}{M_{N 2}} \times 288 K = \frac{32}{28} 288 = 329 K = 56 deg celcius$
Option C is correct.

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The temperature at which the velocity of sound in oxygen will be the same as that of nitrogen at 15∘C is

The correct option is C 56∘CVs=√γRTMMO2=32 , MN2=24VO2at T=VN2at 288 K,T=MO2MN2×288 K=3228288=329 K=56 deg celciusOption C is correct.

The temperature at which the velocity of sound in oxygen will be the same as that of nitrogen at $15^{\circ} C$ is

The correct option is C $56^{\circ} C$
$V_{s} = \sqrt{\frac{γ R T}{M}}$
$M_{O_{2}} = 32$ , $M_{N_{2}} = 24$
$V_{O_{2}} at T = V_{N_{2}} at 288 K$ ,
$T = \frac{M_{O_{2}}}{M_{N 2}} \times 288 K = \frac{32}{28} 288 = 329 K = 56 deg celcius$
Option C is correct.