Yes. All contain the same number of the respective molecules.
No. The root mean square speed of neon is the largest.
Since the three vessels have the same capacity, they have the same volume.
Hence, 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 × 1023
The root mean square speed (\(v_{rms}\)) of a gas of mass m, and temperature T, is given by the relation:
\(v_{rms}=\sqrt\frac{3KT}{m}\)
Where, k is Boltzmann constant
For the given gases, k and T are constants.
Hence v rms depends only on the mass of the atoms, i.e.,
\(v_{rms}=∝\sqrt\frac{1}{m}\)
Therefore, 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.
Hence, neon has the largest root mean square speed among the given gases.
A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s–1 ?
Figure 12.8 shows plot of \(\frac{PV}{T}\) versus P for 1.00×10–3 kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?
(b) Which is true: T1>T2 or T1<T2 ?
(c) What is the value of \(\frac{PV}{T}\) where the curves meet on the y-axis?
(d) If we obtained similar plots for \(1.00×10^–3\) kg of hydrogen, would we get the same value of \(\frac{PV}{T}\) at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of \(\frac{PV}{T}\) (for low pressure high temperature region of the plot) ? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mo1–1 K–1.)