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Helium atom is two times heavier than a hydrogen molecule. At \(298 \mathrm{~K}\), the average kinetic energy of a helium atom is (a) same as that of a hydrogen molecule (b) half that of a hydrogen molecule (c) two times that of a hydrogen molecule (d) four times that of a hydrogen molecule

Short Answer

Expert verified
(a) same as that of a hydrogen molecule

Step by step solution

01

Understand Kinetic Energy Formula

The average kinetic energy of any gas molecule at a given temperature is given by the formula \( KE = \frac{3}{2}kT \), where \( k \) is the Boltzmann constant and \( T \) is the temperature in Kelvin. This formula indicates that the kinetic energy depends only on the temperature, not on the mass of the particle or molecule.
02

Analyze Kinetic Energy at Given Temperature

Considering that both helium atoms and hydrogen molecules are at the same temperature (298 K), their kinetic energies will be calculated using the same formula \( KE = \frac{3}{2}kT \). Since both gases are at the same temperature, their average kinetic energies are therefore equal.
03

Compare Helium and Hydrogen Kinetic Energies

Since we determined in Step 2 that the kinetic energies of helium atoms and hydrogen molecules depend solely on temperature (which is the same for both), the average kinetic energy of a helium atom is the same as that of a hydrogen molecule.

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Key Concepts

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

Boltzmann Constant
The Boltzmann constant, denoted by the symbol \( k \), plays a crucial role in the understanding of kinetic energy in gases. This constant connects the microscopic properties of particles, such as atoms and molecules, to macroscopic properties like temperature. The value of the Boltzmann constant is approximately \( 1.38 \times 10^{-23} \) J/K. It is one of the fundamental constants in physics and helps bridge the gap between macroscopic and microscopic physical laws.
  • It quantifies the relationship between particle kinetic energy and the temperature of their environment.
  • In the context of gases, it shows that the average kinetic energy of a particle is directly proportional to the absolute temperature.
The Boltzmann constant plays a pivotal role in the formula for average kinetic energy, serving as a constant of proportionality that allows us to understand how energy at the particle level translates to observable effects at the macroscopic level.
Temperature Dependence of Kinetic Energy
One of the key insights from gas physics is that the kinetic energy of gas particles depends solely on temperature. This is highlighted by the average kinetic energy equation \( KE = \frac{3}{2}kT \).
  • The kinetic energy does not depend on the type or mass of the gas particles.
  • Instead, it depends entirely on the absolute temperature (measured in Kelvin).
For instance, when both helium atoms and hydrogen molecules are at the same temperature, their average kinetic energies are identical. This fundamental principle is derived from the understanding that temperature is a measure of the average kinetic energy in a substance. Temperature dictates how fast the particles are moving, and therefore, higher temperatures result in higher particle movement, equating to increased kinetic energy. This relationship simplifies many calculations in thermodynamics by focusing only on temperature changes rather than individual molecular characteristics.
Avearge Kinetic Energy Formula
The average kinetic energy of gas particles is given by the formula \( KE = \frac{3}{2}kT \), where \( k \) is the Boltzmann constant and \( T \) is the temperature in Kelvin. This formula is central to thermodynamics and statistical mechanics, providing a measure of the energy due to the motion of particles.
  • The formula shows that kinetic energy is proportional to temperature, reinforcing the idea that particle energy increases with temperature.
  • This relationship is pivotal in understanding various phenomena like gas pressure and diffusion.
The average kinetic energy is consistent across different types of gases at the same temperature, underscoring a universal property of matter in gaseous form. Despite differences in molecular weight between helium and hydrogen, at a constant temperature, helium atoms have the same average kinetic energy as hydrogen molecules because of this formula. This universality simplifies comparative analysis and predictions across chemical and physical processes, exemplifying the power of the kinetic theory of gases.

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Most popular questions from this chapter

A closed container contains equal number of oxygen and hydrogen molecules at a total pressure of \(740 \mathrm{~mm}\). If oxygen is removed from the system then pressure will (a) become double of \(740 \mathrm{~mm}\) (b) become half of \(740 \mathrm{~mm}\) (c) become \(1 / 9\) of \(740 \mathrm{~mm}\) (d) remain unchanged

The values of van der Waals constant ' \(\alpha\) ' for the gases \(\mathrm{O}_{2}, \mathrm{~N}_{2}, \mathrm{NH}_{3}\) and \(\mathrm{CH}_{4}\) are \(1.360,1.390,4.170\) and \(2.253\) L atm. mol \(^{2}\) respectively. The gas which can most easily be liquefied is (a) \(\mathrm{O}_{2}\) (b) \(\mathrm{N}_{2}\) (c) \(\mathrm{NH}_{3}\) (d) \(\mathrm{CH}_{4}\)

In Haber's process, \(30 \mathrm{~L}\) of dihydrogen and \(30 \mathrm{~L}\) of dinitrogen were taken for reaction which yielded only \(50 \%\) of expected product. What is the composition of the gaseous mixture under afore-said conditions in the end? (a) \(20 \mathrm{~L} \mathrm{NH}_{3}, 25 \mathrm{~L} \mathrm{~N}_{2}, 15 \mathrm{~L} \mathrm{H}_{2}\) (b) \(20 \mathrm{~L} \mathrm{NH}_{3}, 20 \mathrm{~L} \mathrm{~N}_{2}, 20 \mathrm{~L} \mathrm{H}_{2}\) (c) \(10 \mathrm{~L} \mathrm{NH}_{3}, 25 \mathrm{~L} \mathrm{~N}_{2}, 15 \mathrm{~L} \mathrm{H}_{2}\) (d) \(20 \mathrm{~L} \mathrm{NH}_{3}, 10 \mathrm{~L} \mathrm{~N}_{2}, 30 \mathrm{~L} \mathrm{H}_{2}\)

If Vrms of \(\mathrm{H}_{2}\) at \(300 \mathrm{~K}\) is \(1.9 \times 10^{3} \mathrm{~m} / \mathrm{s}\). What is the value of Vrms of \(\mathrm{O}_{2}\) at \(1200 \mathrm{~K} ?\) (a) \(1.9 \times 10^{3}\) (b) \(3.8 \times 10^{3} \mathrm{~m} / \mathrm{s}\) (c) \(0.475 \times 10^{3} \mathrm{~m} / \mathrm{s}\) (d) \(0.95 \times 10^{3} \mathrm{~m} / \mathrm{s}\)

Four one litre flasks are separately filled with the gases \(\mathrm{O}_{2}, \mathrm{~F}_{2}, \mathrm{CH}_{4}\) and \(\mathrm{CO}_{2}\) under same conditions. The ratio of the number of molecules in these gases are (a) \(2: 2: 4: 3\) (b) \(1: 1: 1: 1\) (c) \(1: 2 ; 3 ; 4\) (d) \(2: 2 ; 3: 4\)

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