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Chapter 10: Problem 1448

Two monoatomic ideal gases 1 and 2 has molecular weights \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\). Both are kept in two different containers at the same temperature. The ratio of velocity of sound wave in gas 1 and 2 is $\ldots \ldots \ldots$ (A) \(\sqrt{\left(m_{2} / m_{1}\right)}\) (B) \(\sqrt{\left(m_{1} / m_{2}\right)}\) (C) \(\left(\mathrm{m}_{1} / \mathrm{m}_{2}\right)\) (D) \(\left(\mathrm{m}_{2} / \mathrm{m}_{1}\right)\)

Short Answer

Expert verified
The ratio of the velocity of sound wave in gas 1 and 2 is \(\sqrt{\frac{m_{2}}{m_{1}}}\).

Step by step solution

01

Write the formula for the velocity of sound in ideal gases

The formula for the velocity of sound in a monoatomic ideal gas is given by: \(v = \sqrt{\frac{\gamma P}{\rho}}\) Where \(v\) is the velocity of sound, \(\gamma\) is the adiabatic constant (for a monoatomic ideal gas, \(\gamma = \frac{5}{3}\), \(P\) is the pressure of the gas, and \(\rho\) is the density of the gas.
02

Write the density in terms of molecular weight

The density of a gas is related to its molecular weight as follows: \(\rho = \frac{mN}{V}\) Where \(m\) is the molecular weight of the gas, \(N\) is the amount of substance, and \(V\) is the volume of the gas.
03

Substitute the density expression in the formula for the velocity of sound

By substituting the expression for density (\(\rho = \frac{mN}{V}\)) into the formula for the velocity of sound, we have: \(v = \sqrt{\frac{\gamma P}{\frac{mN}{V}}}\)
04

Simplifying the formula for the velocity of sound

Now if we simplify the above formula, we get: \(v = \sqrt{\frac{\gamma PV}{mN}}\)
05

Relate the pressure, volume, amount of substance, and temperature for ideal gases

According to the ideal gas law: \(PV = NRT\) Where \(R\) is the gas constant, and \(T\) is the temperature.
06

Find the expressions for the velocities of sound in both gases

For the gases 1 and 2 at the same temperature, the ideal gas law becomes: \(P_1V_1 = N_1RT\) \(P_2V_2 = N_2RT\) Now, we can find the expressions for the velocities of sound in both gases using these equations and the simplified formula for the velocity of sound: \(v_1 = \sqrt{\frac{\frac{5}{3}P_1V_1}{m_1N_1}} = \sqrt{\frac{\frac{5}{3}N_1RT}{m_1N_1}} = \sqrt{\frac{5R}{3m_1}T}\) Similarly for gas 2, \(v_2 = \sqrt{\frac{5R}{3m_2}T}\)
07

Find the ratio of velocities

To find the ratio of velocities of sound in gas 1 and 2, divide \(v_1\) by \(v_2\): \(\frac{v_1}{v_2} = \frac{\sqrt{\frac{5R}{3m_1}T}}{\sqrt{\frac{5R}{3m_2}T}} = \sqrt{\frac{\frac{5R}{3m_1}T}{\frac{5R}{3m_2}T}} = \sqrt{\frac{m_2}{m_1}}\) The ratio is equal to \(\sqrt{\frac{m_2}{m_1}}\). Therefore, the correct answer is (A).

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Most popular questions from this chapter

A body of mass \(1 \mathrm{~kg}\) suspended from the free end of a spring having force constant \(400 \mathrm{Nm}^{-1}\) is executing S.H.M. When the total energy of the system is 2 joule, the maximum acceleration is $\ldots \ldots . \mathrm{ms}^{-2}$. (A) \(8 \mathrm{~ms}^{-2}\) (B) \(10 \mathrm{~ms}^{-2}\) (C) \(40 \mathrm{~ms}^{-2}\) (D) \(40 \mathrm{cms}^{-2}\)

A block having mass \(\mathrm{M}\) is placed on a horizontal frictionless surface. This mass is attached to one end of a spring having force constant \(\mathrm{k}\). The other end of the spring is attached to a rigid wall. This system consisting of spring and mass \(\mathrm{M}\) is executing SHM with amplitude \(\mathrm{A}\) and frequency \(\mathrm{f}\). When the block is passing through the mid-point of its path of motion, a body of mass \(\mathrm{m}\) is placed on top of it, as a result of which its amplitude and frequency changes to \(\mathrm{A}^{\prime}\) and \(\mathrm{f}\). The ratio of amplitudes $\left(\mathrm{A}^{1} / \mathrm{A}\right)=\ldots \ldots \ldots$ (A) \(\sqrt{\\{}(\mathrm{M}+\mathrm{m}) / \mathrm{m}\\}\) (B) \(\sqrt{\\{m} /(\mathrm{M}+\mathrm{m})\\}\) (C) \(\sqrt{\\{} \mathrm{M} /(\mathrm{M}+\mathrm{m})\\}\) (D) \(\sqrt{\\{}(\mathrm{M}+\mathrm{m}) / \mathrm{M}\\}\)

The ratio of frequencies of two waves travelling through the same medium is \(2: 5 .\) The ratio of their wavelengths will be.... (A) \(2: 5\) (B) \(5: 2\) (C) \(3: 5\) (D) \(5: 3\)

Two simple pendulums having lengths \(144 \mathrm{~cm}\) and \(121 \mathrm{~cm}\) starts executing oscillations. At some time, both bobs of the pendulum are at the equilibrium positions and in same phase. After how many oscillations of the shorter pendulum will both the bob's pass through the equilibrium position and will have same phase ? (A) 11 (B) 12 (C) 21 (D) 20

A particle is executing S.H.M. between \(\mathrm{x}=-\mathrm{A}\) and \(\mathrm{x}=+\mathrm{A}\). If the time taken by the particle to travel from \(\mathrm{x}=0\) to \(\mathrm{A} / 2\) is \(\mathrm{T}_{1}\) and that taken to travel from \(\mathrm{x}=\mathrm{A} / 2\) to \(\mathrm{x}=\mathrm{A}\) is \(\mathrm{T}_{2}=\) then \(\ldots .\) (A) \(\mathrm{T}_{1}<\mathrm{T}_{2}\) (B) \(\mathrm{T}_{1}>\mathrm{T}_{2}\) (C) \(\mathrm{T}_{1}=2 \mathrm{~T}_{2}\) (D) \(\mathrm{T}_{1}=\mathrm{T}_{2}\)
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