Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is V_{rms} the largest?

Asked by Pragya Singh | 1 year ago | 88

All the three vessels have the same capacity, they have the same volume.

So, each gas has the same pressure, volume and temperature

According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules.

This number is equal to Avogadro’s number, N = 6.023 x 10^{23}.

The root mean square speed (V_{rms}) of a gas of mass m and temperature T is given by the relation:

V_{rms} = \( \sqrt{\dfrac{3KT}{M}}\)

Where,

k is Boltzmann constant

For the given gases, k and T are constants

Therefore, V_{rms} depends only on the mass of the atoms, i.e., V_{rms} ∝\((\dfrac{1}{M})^\dfrac{1}{2}\)

Hence, the root mean square speed of the molecules in the three cases is not the same.

Among neon, chlorine and uranium hexafluoride, the mass of neon is the smallest.

Therefore, neon has the largest root mean square speed among the given gases.

Answered by Abhisek | 1 year agoGiven below are densities of some solids and liquids. Give rough estimates of the size of their atoms :

Substance |
Atomic Mass (u) |
Kg m^{-3}) |

Carbon (diamond) | 12.01 | 2.22 |

Gold | 197.00 | 19.32 |

Nitrogen (liquid) | 14.01 | 1.00 |

Lithium | 6.94 | 0.53 |

Fluorine (liquid) | 19.00 | 1.14 |

[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].

A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have a uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres n_{2}= n_{1} exp [ -mg \(
\dfrac{(h_2–h_1)}{k_BT}\)T] where n_{2}, n_{1 }refer to number density at heights h_{2 }and h_{1 }respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:n_{2} = n_{1} exp [ -mg N_{A} (ρ – ρ′ ) \(
\dfrac{(h_2–h_1)}{ρ RT}\)] where ρ is the density of the suspended particle, and ρ′ ,that of surrounding medium. [N_{A} is Avogadro’s number, and R the universal gas constant.] [Hint : Use Archimedes principle to find the apparent weight of the suspended particle.]

From a certain apparatus, the diffusion rate of hydrogen has an average value of \( 28.7 cm^3 s^{-1}\). The diffusion of another gas under the same conditions is measured to have an average rate of \( 7.2 cm^3 s^{-1}\). Identify the gas. [Hint: Use Graham’s law of diffusion: \( \dfrac{R_1}{R_2}\) = \( (\dfrac{M_2}{M_1})^ \dfrac{1}{2}\), where \( R_1 , R_2\) are diffusion rates of gases 1 and 2, and \( M_1\) and \( M_2\) their respective molecular masses. The law is a simple consequence of kinetic theory.]

A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 170°C . Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of \( N_2\) = 28.0 u).