/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 A gas is confined inside a conta... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 35

A gas is confined inside a container having a movable piston. The gas is allowed to expand isobarically. If the initial volume of gas is \(V_{0}\) and the speed of sound in the gas is \(C_{0}\), then the speed of sound when the volume of the gas increases to \(4 V_{0}\) is (A) \(C_{0}\) (B) \(2 C_{0}\) (C) \(4 C_{0}\) (D) \(C_{0} / 2\)

Short Answer

Expert verified
The speed of sound when the volume of the gas increases to \(4V_{0}\) is \(2C_{0}\), which corresponds to option (B).

Step by step solution

01

Write down the speed of sound equation

First, we need to write down the formula for the speed of sound in a gas. The equation for the speed of sound in a gas is given as: \[C = \sqrt{\gamma \frac{P}{\rho}}\] where \(C\) is the speed of sound, \(\gamma\) is the adiabatic index (ratio of specific heats), \(P\) is the pressure, and \(\rho\) is the density of the gas.
02

Write down the isobaric process relationship for density and volume

Since the expansion is isobaric, the pressure remains constant. Thus, we can use the following relationship between initial and final density and volume: \[\frac{\rho_{0}}{\rho} = \frac{V}{V_{0}}\] where \(\rho_{0}\) is the initial density, \(\rho\) is the final density, and \(V\) is the final volume.
03

Substitute the isobaric relationship into the speed of sound equation

By substituting the isobaric relationship into the speed of sound equation, we get: \[C = \sqrt{\gamma \frac{P}{\rho_{0}} V_{0} \frac{V}{V_{0}}}\] Simplifying, we have: \[C = \sqrt{\gamma \frac{P}{\rho_{0}}V}\]
04

Find the initial speed of sound

The initial speed of sound, \(C_{0}\), can be found by substituting the initial volume (\(V_{0}\)): \[C_{0} = \sqrt{\gamma \frac{P}{\rho_{0}}V_{0}}\]
05

Find the speed of sound at the final volume

Now, we need to find the speed of sound when the volume of gas increases to \(4V_{0}\). We'll substitute the final volume into the speed of sound equation: \[C = \sqrt{\gamma \frac{P}{\rho_{0}}(4V_{0})}\]
06

Simplify and solve for the final speed of sound

Simplifying the equation above, we get: \[C = \sqrt{4\gamma \frac{P}{\rho_{0}}V_{0}} = \sqrt{4C_{0}^{2}}\] Taking the square root of both sides: \[C = 2C_{0}\] So, the speed of sound when the volume of the gas increases to \(4V_{0}\) is \(2C_{0}\), which corresponds to option (B).

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Most popular questions from this chapter

Two moles of an ideal gas at \(300 \mathrm{~K}\) were cooled at constant volume so that the pressure is reduced to half the initial value. As a result of heating at constant pressure, the gas has expanded till it attains the original temperature. The total heat absorbed by gas, if \(R\) is the gas constant (A) \(150 R\) (B) \(300 R\) (C) \(75 R\) (D) \(100 R\)

One mole of an ideal gas is enclosed in a cylinder fitted with a frictionless piston and occupies a volume of \(1.5\) litre at a pressure of \(1.2 \mathrm{~atm} .\) It is subjected to a process given by equation \(T=\alpha V^{2}, \gamma\) (adiabatic constant) for the gas \(=1.5\). Choose the wrong statement. Given \(R \alpha=80 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{lit}^{-2}(R=\) gas constant and \(\alpha\) is constant) (A) The \(P-V\) diagram of the process is a straight line. (B) The work done by the gas in increasing the volume of the gas to 9 litre is \(3150 \mathrm{~J}\). (C) The change in the internal energy of the gas is \(12600 \mathrm{~J}\). (D) The heat supplied to the gas in the process is 1575 J.

The molar specific heats of an ideal gas at constant pressure and volume are denoted by \(C_{p}\) and \(C_{v}\), respectively. Further, \(\frac{C_{p}}{C_{v}}=\gamma\) and \(R\) is the gas constant for 1 gm mole of a gas. Then \(C_{v}\) is equal to (A) \(R\) (B) \(\gamma R\) (C) \(\frac{R}{\gamma-1}\) (D) \(\frac{\gamma R}{\gamma-1}\)

Equal amount of same gas in two similar cylinders, \(A\) and \(B\), compressed to same final volume from same initial volume one adiabatically and another isothermally, respectively, then (A) final pressure in \(A\) is more than in \(B\). (B) final pressure in \(B\) is greater than in \(A\). (C) final pressure in both equal. (D) for the gas, value of \(\gamma=\frac{C_{p}}{C_{V}}\) is required.

Consider two containers \(A\) and \(B\) containing identical gases at the same pressure, volume, and temperature. The gas in container \(A\) is compressed to half of its original volume isothermally while the gas in container \(B\) is compressed to half of its original value adiabatically. The ratio of final pressure of gas in \(B\) to that of gas in \(A\) is (A) \(2^{\gamma-1}\) (B) \(\left(\frac{1}{2}\right)^{\gamma-1}\) (C) \(\left(\frac{1}{1-\gamma}\right)^{2}\) (D) \(\left(\frac{1}{\gamma-1}\right)^{2}\)
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