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Physics Question on Waves

According to Newtons formula, the speed of sound in air at STPSTP is (Take the mass of 11 mole of air is 29×10−3 kg29 \times 10^{-3}\, kg)

A

250 ms−1250\, ms^{-1}

B

260 ms−1260\, ms^{-1}

C

270 ms−1270\, ms^{-1}

D

280 ms−1280\, ms^{-1}

Answer

280 ms−1280\, ms^{-1}

Explanation

Solution

11 mole of any gas occupies 22.422.4 litres at STPSTP.bnm Therefore, density of air at STPSTP is ρ=Mass of one mole of airVolume of one mole of air at STP\rho = \frac{\text{Mass of one mole of air}}{\text{Volume of one mole of air at }STP} =29×10−3 kg22.4×10−3 m3=1.29 kg m3= \frac{29 \times 10^{-3}\,kg}{22.4 \times 10^{-3}\,m^{3}} = 1.29\,kg\,m^{3} At STPSTP, P=1 atm=1.01×105 N m−2P = 1\, atm = 1.01 \times 10^{5}\, N \,m^{-2} According to Newtons formula, the speed of sound in air at STPSTP is v=(pρ)=1.01×105 N m−229×kg m−3v = \sqrt{\left(\frac{p}{\rho}\right)} = \sqrt{\frac{1.01 \times 10^{5}\, N \,m^{-2}}{29 \times kg\,m^{-3}}} =280 ms−3= 280\,ms^{-3}