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By how may folds the temperature of the gas would increase when the RMS velocity of gas molecules in a container of fixed volume is increased from \(5 \times 10^{4} \mathrm{~cm}\) \(\sec ^{-1}\) to \(10 \times 10^{4} \mathrm{~cm} \sec -1 ?\) (a) 6 times (b) 4 times (c) 2 times (d) 8 times

Short Answer

Expert verified
The temperature of the gas increases 4 times, option (b).

Step by step solution

01

Understanding RMS Velocity and Temperature Relationship

The root-mean-square (RMS) velocity of gas molecules is directly related to the temperature of the gas. In physics, the formula for RMS velocity \( v \) is \( v = \sqrt{\frac{3kT}{m}} \), where \( k \) is Boltzmann's constant, \( T \) is temperature, and \( m \) is the mass of a gas molecule.
02

Setting Up the Ratio of RMS Velocities

We have an initial and a final RMS velocity. Let \( v_1 = 5 \times 10^4 \) cm/s and \( v_2 = 10 \times 10^4 \) cm/s. We need to find the ratio of \( \frac{v_2}{v_1} \). Calculating gives us: \( \frac{v_2}{v_1} = \frac{10 \times 10^4}{5 \times 10^4} = 2 \).
03

Relating the Ratio to Temperature

Since RMS velocity is proportional to the square root of temperature \( (v \propto \sqrt{T}) \), if \( \frac{v_2}{v_1} = 2 \), then \( \frac{\sqrt{T_2}}{\sqrt{T_1}} = 2 \).
04

Calculating the Temperature Ratio

Square both sides of \( \frac{\sqrt{T_2}}{\sqrt{T_1}} = 2 \) to find \( \frac{T_2}{T_1} = 2^2 = 4 \). The temperature increases by this factor.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Temperature and RMS velocity relationship
The relationship between temperature and the root-mean-square (RMS) velocity of gas molecules is a foundational concept in kinetic molecular theory. Essentially, the RMS velocity is a direct function of the temperature of the gas. The formula that describes this relationship is given by:
  • \( v = \sqrt{\frac{3kT}{m}} \)
Here, \( v \) stands for the RMS velocity, \( k \) is Boltzmann's constant, \( T \) represents temperature, and \( m \) is the mass of a single molecule of the gas. The equation tells us that as the temperature of the gas increases, the speed at which the molecules move also increases.
RMS velocity is indicative of the average speed of gas molecules within a substance at a given temperature. As a result, there is a proportional relationship between the RMS velocity of the molecules and the square root of the temperature. Higher temperatures mean that molecules move faster because they have more kinetic energy. This is crucial in understanding the behavior of gases under different conditions.
Ratio of RMS velocities
When comparing two different states of a gas, it's useful to understand how changes in RMS velocity reflect changes in temperature. In the original exercise, the RMS velocity is initially \(5 \times 10^4 \text{ cm/s}\), and later becomes \(10 \times 10^4 \text{ cm/s}\). To understand this change, we calculate the ratio of these velocities:
  • \( \frac{v_2}{v_1} = \frac{10 \times 10^4}{5 \times 10^4} = 2 \)
This calculation tells us that the RMS velocity doubled. Since RMS velocity is proportional to the square root of the temperature (\( v \propto \sqrt{T} \)), it follows that if the velocity increases by a factor of 2, the square roots of their corresponding temperatures have also doubled.
To find how much the temperature itself changes, square the factor of change in velocity:
  • \( \frac{T_2}{T_1} = (\frac{v_2}{v_1})^2 = 2^2 = 4 \)
Thus, the temperature of the gas increases by a factor of 4.
Boltzmann's constant
Boltzmann's constant (denoted as \( k \)) is a fundamental constant in physics that plays a vital role in the study of gases. It links the macroscopic and microscopic worlds by relating individual particle movements to macroscopic properties like temperature and energy. The value of Boltzmann's constant is approximately \( 1.38 \times 10^{-23} \text{ J/K} \).
The role of Boltzmann's constant is especially clear in the formula for RMS velocity \( v = \sqrt{\frac{3kT}{m}} \). Here, it scales the temperature (\( T \)) to energy terms within the ideal gas context. Essentially, it helps us understand how thermal energy translates to the movement of individual molecules.
Using Boltzmann's constant allows us to accurately calculate particle speed and energy, thereby highlighting its importance in thermodynamics and statistical mechanics. It's a key ingredient for deriving expressions that connect molecular velocity and temperature, like the RMS velocity formula, affirming its role in interpreting molecular-scale phenomena.

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Most popular questions from this chapter

At \(27^{\circ} \mathrm{C}\), a closed vessel contains a mixture of equal weights of helium (mol. wt \(=4\) ), methane (mol. wt \(=16\) ) and sulphur dioxide (mol. wt \(=64\) ). The pressure exerted by the mixture is \(210 \mathrm{~mm}\). If the partial pressures of helium, methane and sulphur dioxide are \(\mathrm{P}_{1}, \mathrm{P}_{2}\) and \(\mathrm{P}_{3}\) respectively, which one of the following is correct? (a) \(\mathrm{P}_{3}>\mathrm{P}_{2}>\mathrm{P}_{1}\) (b) \(\mathrm{P}_{1}>\mathrm{P}_{2}>\mathrm{P}_{3}\) (c) \(\mathrm{P}_{1}>\mathrm{P}_{3}>\mathrm{P}_{2}\) (d) \(\mathrm{P}_{2}>\mathrm{P}_{3}>\mathrm{P}_{1}\)

At a certain pressure, volume of a gas at \(27^{\circ} \mathrm{C}\) is 20 litre. If the pressure and temperature are doubled, its volume will be (a) 20 litre (b) 40 litre (c) \(8.2\) litre (d) \(10.9\) litre

If \(\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3} \ldots \ldots \ldots\) represents the speed of \(\mathrm{n}_{1}\), \(\mathrm{n}_{2}, \mathrm{n}_{3}, \ldots\) molecules, then the root mean square of speed is (a) \(\left(\frac{\mathrm{n}_{1} \mathrm{C}_{1}^{2}+\mathrm{n}_{2} \mathrm{C}_{2}^{2}+\mathrm{n}_{3} \mathrm{C}_{3}^{2}+\ldots}{\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}+\ldots}\right)^{1 / 2}\) (b) \(\left(\frac{n_{1} C_{1}^{2}+n_{2} C_{2}^{2}+n_{3} C_{3}^{2}+\ldots}{n_{1}+n_{2}+n_{3}+\ldots}\right)^{2}\) (c) \(\frac{\left(\mathrm{n}_{1} \mathrm{C}_{1}^{2}\right)^{1 / 2}}{\mathrm{n}_{1}}+\frac{\left(\mathrm{n}_{2} \mathrm{C}_{2}^{2}\right)^{1 / 2}}{\mathrm{n}_{2}}+\frac{\left(\mathrm{n}_{3} \mathrm{C}_{3}^{2}\right)^{1 / 2}}{\mathrm{n}_{3}}+\ldots\) (d) \(\left[\frac{\left(\mathrm{n}_{1} \mathrm{C}_{1}+\mathrm{n}_{2} \mathrm{C}_{2}+\mathrm{n}_{3} \mathrm{C}_{3}+\ldots\right)^{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}+\ldots}\right]^{1 / 2}\)

The density of neon will be highest at (a) STP (b) \(0^{\circ} \mathrm{C}, 2 \mathrm{~atm}\) (c) \(273^{\circ} \mathrm{C}, 1 \mathrm{~atm}\) (d) \(273^{\circ} \mathrm{C}, 2 \mathrm{~atm}\)

\(\mathrm{A}\) and \(\mathrm{B}\) are ideal gases. The molecular weights of \(\mathrm{A}\) and \(\mathrm{B}\) are in the ratio of \(1: 4\). The pressure of a gas mixture containing equal weights of \(\mathrm{A}\) and \(\mathrm{B}\) is \(\mathrm{P}\) atm. What is the partial pressure (in atm) of B in the mixture? (a) \(\mathrm{P} / 5\) (b) \(\mathrm{P} / 2\) (c) \(\mathrm{P} / 2.5\) (d) \(3 \mathrm{P} / 4\)

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