Do all the molecules in a 1 -mol sample of \(\mathrm{CH}_{4}(g)\) have the same kinetic energy at \(273 \mathrm{~K}\) ? Do all molecules in a \(1-\mathrm{mol}\) sample of \(\mathrm{N}_{2}(g)\) have the same velocity at \(546 \mathrm{~K}\) ? Explain.

Short Answer

Expert verified
In a 1-mol sample of CH4(g), not all molecules have the same kinetic energy at 273 K, although their average kinetic energy is the same due to the temperature. Similarly, in a 1-mol sample of N2(g), not all molecules have the same velocity at 546 K, despite their average kinetic energy being the same due to the temperature. This variation in kinetic energy and velocity among individual molecules occurs because of their unique motion and collisions within the gas sample.

Step by step solution

01

Understand the relationship between temperature and kinetic energy

The average kinetic energy of a molecule in a gas sample is directly proportional to the temperature, meaning that increasing the temperature will increase the average kinetic energy of the molecules. The formula to find the average kinetic energy of a molecule can be written as: \[K.E = \dfrac{3}{2} kT\] Where K.E is the average kinetic energy, k is the Boltzmann constant, and T is the temperature in Kelvin. It is essential to note that this formula provides us with the average kinetic energy of the molecules in a sample, not the kinetic energy of each molecule.
02

Kinetic energy of CH4 molecules at 273 K

Now, we know that the average kinetic energy of the sample depends on the temperature. However, each molecule's kinetic energy can be different due to their unique velocities and collisions that occur within the gas sample. Even though the average kinetic energy at 273 K can be the same for all molecules in a 1-mol sample of CH4(g), individual molecules can have a range of different kinetic energies.
03

Relationship between velocity, mass, and kinetic energy

The kinetic energy of a molecule is related to its mass and velocity by the following formula: \[K.E = \dfrac{1}{2} mv^2\] Where K.E is the kinetic energy, m is the mass of the gas molecule, and v is its velocity. In this exercise, we are interested in examining the individual molecular velocities of N2(g) at 546 K.
04

Velocity of N2 molecules at 546 K

It is important to remember that molecular velocity, temperature, mass, and kinetic energy are related. Therefore, even if the average kinetic energy of the N2 molecules at 546 K is the same, their individual velocities can vary because of their unique pathway through the sample and collisions with other gas molecules. Hence, all molecules in a 1-mol sample of N2 do not have the same velocity at 546 K. In conclusion, - All molecules in a 1-mol sample of CH4(g) do not have the same kinetic energy at 273 K, even though their average kinetic energy is the same due to the temperature. - All molecules in a 1-mol sample of N2(g) do not have the same velocity at 546 K, even though their average kinetic energy is the same due to the temperature.

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Most popular questions from this chapter

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One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming \(\mathrm{Be}^{3+}\) ions) and that it gave an oxide with the formula \(\mathrm{Be}_{2} \mathrm{O}_{3} .\) This resulted in a calculated atomic mass of \(13.5\) for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming \(\mathrm{Be}^{2+}\) ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of \(9.0 .\) In 1894 , A. Combes (Comptes Rendus 1894, p. 1221\()\) reacted beryllium with the anion \(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}^{-}\) and measured the density of the gaseous product. Combes's data for two different experiments are as follows: If beryllium is a divalent metal, the molecular formula of the product will be \(\mathrm{Be}\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{2} ;\) if it is trivalent, the formula will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{3}\). Show how Combes's data help to confirm that beryllium is a divalent metal.

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