Determine the speed of sound in oxygen using the following data. At STP ( T = 273 K and p = 1.0 × 10^5 N m^ -2 ), the mass of 22.4 litre of oxygen is 32 g . The molar heat capacity of oxygen is C_V = 2.5 R at constant

Correct answer: A. 313 m s^ -1 The ratio of specific heats is γ = (C_p)/(C_V) = (3.5 R)/(2.5 R) = 1.4 The density of oxygen at STP is = (M)/(V) = 32 × 10^ -3 kg 22.4 × 10^ -3 m^3 = (10)/(7) kg m^ -3 The speed of sound in a gas is given by Laplace's formula v = (γ p)/( ) Substituting the given values: v = (1.4 × 1.0 × 10^5)/(10/7) v = 1.4 × 0.7 × 10^5 v = 0.98 × 10^5 = 98000 v 313 m s^ -1

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Sound Waves

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Sound Waves

Determine the speed of sound in oxygen using the following data. At...

Determine the speed of sound in oxygen using the following data. At STP (

T = 273 \text{ K}

and

p = 1.0 \times 10^5 \text{ N m}^{-2}

), the mass of

22.4

litre of oxygen is

32 \text{ g}

. The molar heat capacity of oxygen is

C_V = 2.5 R

at constant volume and

C_p = 3.5 R

at constant pressure.

Options

A $313 \text{ m s}^{-1}$
B $224 \text{ m s}^{-1}$
C $342 \text{ m s}^{-1}$
D $265 \text{ m s}^{-1}$

Correct answer

A. $313 \text{ m s}^{-1}$

Step-by-step solution

The ratio of specific heats is

\gamma = \dfrac{C_p}{C_V} = \dfrac{3.5 R}{2.5 R} = 1.4

The density of oxygen at STP is

\rho = \dfrac{M}{V} = \dfrac{32 \times 10^{-3} \text{ kg}}{22.4 \times 10^{-3} \text{ m}^3} = \dfrac{10}{7} \text{ kg m}^{-3}

The speed of sound in a gas is given by Laplace's formula

v = \sqrt{\dfrac{\gamma p}{\rho}}

Substituting the given values:

v = \sqrt{\dfrac{1.4 \times 1.0 \times 10^5}{10/7}}

v = \sqrt{1.4 \times 0.7 \times 10^5}

v = \sqrt{0.98 \times 10^5} = \sqrt{98000}

v \approx 313 \text{ m s}^{-1}

Practice Sound Waves on Quantrex Academy →

More from Sound Waves