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If the rms velocity of a gas at \(100 \mathrm{~K}\) is \(10^{4} \mathrm{~cm} \mathrm{sec}^{-1}\), what is the temperature (in \({ }^{\circ} \mathrm{C}\) ) at which the rms velocity will be \(3 \times 10^{4} \mathrm{~cm} \sec ^{-1} ?\) (a) 900 (b) 627 (c) 327 (d) 1217

Short Answer

Expert verified
The temperature is 627°C.

Step by step solution

01

Understanding the Concept of rms Velocity

Root mean square (rms) velocity of gas molecules is given by the equation \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( k \) is the Boltzmann constant, \( T \) is the absolute temperature, and \( m \) is the mass of a molecule.
02

Using the Ratio Form of the rms Velocity Equation

From the formula \( v_{rms} \propto \sqrt{T} \), we deduce that \( \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} \). Here \( v_1 = 10^4 \text{ cm/sec} \) and \( v_2 = 3 \times 10^4 \text{ cm/sec} \). Therefore, \( \frac{10^4}{3 \times 10^4} = \sqrt{\frac{100}{T_2}} \).
03

Solving for the New Temperature

First, simplify the equation: \( \frac{1}{3} = \sqrt{\frac{100}{T_2}} \). Squaring both sides gives \( \left(\frac{1}{3}\right)^2 = \frac{100}{T_2} \). This simplifies to \( \frac{1}{9} = \frac{100}{T_2} \). Rearranging gives \( T_2 = 900 \text{ K} \).
04

Converting the Temperature to Celsius

Convert the temperature from Kelvin to Celsius using the formula \( \text{Celsius} = \text{Kelvin} - 273.15 \). Thus, the temperature is \( 900 - 273.15 = 626.85 \approx 627^\circ\text{C} \).

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Key Concepts

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

Root Mean Square Velocity
The root mean square (rms) velocity is a concept that helps us understand the motion of gas particles. It's essentially the square root of the average of the squares of the velocities of the gas particles. This measurement provides a useful average speed for gas particles, which is important in thermodynamics and physical chemistry.

The formula for rms velocity is given by:
  • \( v_{rms} = \sqrt{\frac{3kT}{m}} \)
Here:
  • \( k \) is the Boltzmann constant, which relates to the energy at the molecular level.
  • \( T \) is the temperature in Kelvin, reflecting the average kinetic energy of the particles.
  • \( m \) is the mass of one particle of the gas.
The rms velocity is proportional to the square root of the temperature. As the temperature increases, the rms velocity increases, meaning gas particles move faster. This relationship is central to understanding gas behavior in different temperature conditions.
Temperature Conversion
Temperature conversion is essential for comparing temperatures in different units. Often, scientific calculations use the Kelvin scale, mainly because it's an absolute temperature scale. But, in many everyday contexts, Celsius is used. The conversion between these two is straightforward:
  • \( \text{Celsius} = \text{Kelvin} - 273.15 \)
This tells us that to convert a temperature from Kelvin to Celsius, we simply subtract 273.15.

For instance, if a gas temperature is 900 K, in Celsius this would be:
  • \(900 - 273.15 = 626.85 \approx 627^{\circ}\text{C}\)
This conversion helps in practical applications and problems, especially in chemistry and physics, where measurements might initially be in Kelvin but are often communicated in Celsius for a wider understanding.
Ideal Gas Law
The ideal gas law is a fundamental principle that relates the temperature, pressure, and volume of an ideal gas. Though not directly used in the rms velocity calculation, understanding this law gives insight into gas behavior. The ideal gas law is expressed as:
  • \( PV = nRT \)
In this equation:
  • \( P \) is the pressure, \( V \) is the volume.
  • \( n \) is the number of moles of the gas.
  • \( R \) is the ideal gas constant.
  • \( T \) is the temperature in Kelvin.
This law illustrates how changes in temperature affect the pressure and volume of a gas. For example: increasing temperature will result in increased pressure if the volume of the gas is constrained.

Understanding the ideal gas law helps connect the rms velocity to a broader context of gas behaviors under different conditions.
Boltzmann Constant
The Boltzmann constant, denoted as \( k \), is a fundamental constant in physics and chemistry. It provides a bridge between microscopic and macroscopic physical quantities. The role of the Boltzmann constant is to relate the average kinetic energy of particles in a gas with the temperature of the gas.
  • Its value is approximately \( 1.38 \times 10^{-23} \text{J/K} \).
This makes \( k \) critical for calculations involving molecular energies and various thermodynamic processes.

When using the rms velocity formula \( v_{rms} = \sqrt{\frac{3kT}{m}} \), the Boltzmann constant helps to quantify the relationship between temperature and kinetic energy. Higher temperature means higher energy and hence faster particle speeds. This constant is indispensable for understanding how variations in temperature influence the movement and energy of gas molecules.

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Most popular questions from this chapter

The density of gas \(\mathrm{A}\) is twice of that of \(\mathrm{B}\) and mol. wt. of \(\mathrm{A}\) is half of that of \(\mathrm{B}\). The Ratio of partial pressures of \(P_{A}\) and \(P_{n}\) is (a) \(\frac{1}{4}\) (b) \(\frac{4}{1}\) (c) \(\frac{2}{1}\) (d) \(\frac{1}{2}\)

At constant volume and temperature conditions, the rates of diffusion \(\mathrm{D}_{A}\) and \(\mathrm{D}_{\mathrm{B}}\) of gases \(\mathrm{A}\) and \(\mathrm{B}\) having densities \(\rho_{A}\) and \(\rho_{n}\) are related by the expression (a) \(\mathrm{D}_{\mathrm{A}}=\left[\mathrm{D}_{\mathrm{a}} \rho_{\mathrm{A}} / \rho_{\mathrm{h}}\right]^{1 / 2}\) (b) \(\mathrm{D}_{\mathrm{A}}=\left[\mathrm{D}_{\mathrm{B}} \rho_{\mathrm{B}} / \rho_{A}\right]^{1 / 2}\) (c) \(D_{A}=D_{B}\left[\rho_{A} / \rho_{B}\right]^{1 / 2}\) (d) \(\mathrm{D}_{\mathrm{A}}=\mathrm{D}_{\mathrm{B}}\left[\rho_{\mathrm{B}} / \rho_{\mathrm{A}}\right]^{12}\)

The temperature below which a gas can be made to liquefy by variation in pressure is called its (a) inversion temperature (b) critical temperature (c) neutral temperature (d) curie point

An L.PG. cylinder contains \(15 \mathrm{~kg}\) of butane gas at \(27^{\circ} \mathrm{C}\) and 10 atmospheric pressure. It was leaking and its pressure fell down to 8 atmospheric pressure after one day. The gas leaked in 5 days is (a) \(10 \mathrm{~kg}\) (b) \(3 \mathrm{~kg}\) (c) \(15 \mathrm{~kg}\) (d) \(12 \mathrm{~kg}\)

A gas diffuses four times as quickly as oxygen. The molar weight of gas is (a) 2 (b) 4 (c) 8 (d) 16

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