The mass of 1 mole of air is 29 × 10-3 kg. The speed of sound in air at standard temperature and pressure is (. Take .γ=(7/5))

Question

The mass of 1 mole of air is 29 × 10-3 kg. The speed of sound in air at standard temperature and pressure is (. Take .&gamma;=(7/5)) (A) 270 ms -1 (B) 290 ms -1 (C) 330 ms -1 (D) 350 ms -1

Tardigrade Admin · Accepted Answer

Speed =√(&gamma; P/ρ) ρ=( text Mass of 1 text mole of air / text Volume of 1 text mole at STP ) =(29 × 10-3 kg /22.4 × 10-3 m 3) ρ=1.3 P text Standard =1.01 × 105 Nm -2 ⇒ Speed =√(7 × 1.01 × 105/5 × 1.3) ≃ 330 ms -1 .

Q. The mass of 1 mole of air is $29 \times 1 0^{- 3} k g$ . The speed of sound in air at standard temperature and pressure is $($ Take $γ = \frac{7}{5})$

$270 m s^{- 1}$

$290 m s^{- 1}$

$330 m s^{- 1}$

$350 m s^{- 1}$

Q. The mass of 1 mole of air is 29×10−3kg. The speed of sound in air at standard temperature and pressure is ( Take γ=57​)

270ms−1

290ms−1

330ms−1

350ms−1

Speed =ργP​​ ρ= Volume of 1 mole at STP Mass of 1 mole of air ​ =22.4×10−3m329×10−3kg​ ρ=1.3PStandard ​=1.01×105Nm−2 ⇒ Speed =5×1.37×1.01×105​​ ≃330ms−1.

Q. The mass of 1 mole of air is $29 \times 1 0^{- 3} k g$ . The speed of sound in air at standard temperature and pressure is $($ Take $γ = \frac{7}{5})$

$270 m s^{- 1}$

$290 m s^{- 1}$

$330 m s^{- 1}$

$350 m s^{- 1}$