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Both the sample will contain the same number of molecules, in accordance with Avogadro's law.

Under the initial conditions,

V = 30 litre = 30 x 10^-3 m³

p = 15 atm = 15 x 1.01 x 10⁵ Pa

T = 27°C = 273 + 27 = 300 K

Also; R = 8.31 J mol^-1 K^-1 and molar mass, M = 32 x 10^-3 kg

Using the relation,

pV = uRT

⇒ u = pV/RT = {15 x 1.01 x 10⁵ x 30 x 10^-3}/{8.31 x 300} = 18.23

Now, u = m/M

⇒ m = Mu = 18.23 x 32 x 10^-3 kg

or, m = 0.58 kg

Under the final conditions:

V' = 30 litre = 30 x 10^-3 m³

p' = 11 atm = 11 x 1.01 x 10⁵ Pa

T' = 17°C = 17 + 273 = 290 K

We have, u' = p"V'/RT' = {11 x 1.01 x 10⁵ x 30 x 10^-3}/{8.31 x 290} = 13.83

m' = u'm = 13.83 x 32 x 10^-3 kg = 0.44 kg

Mass of the oxygen taken out of the cylinder

= m - m' = 0.58 - 0.44 = 0.14 kg

(a) The dotted plot shows the pV/T is a constant quantity (= uR)

This signifies the ideal gas behaviour.

(b) Here T₁ > T₂

(c) At the point where the curve meets the y-axis, we have pV/T = uR

where u is the number of moles of oxygen gas. Here, Mass of oxygen,

Also, molecular mass, M = 32 x 10^-3 kg

No. of moles, u = m/M = {1.00 x 10^-3}/{32 x 10^-3} = 1/32

pV/T = uR = 1/32 x 8.31

= 0.26 J K^-1

(d) Since the value of pV/T depends upon the number of moles, we will not get the same value for pV/T in case of hydrogen. To obtain the same value of u (i.e., 1/32) and hence pV/T, we must have

= {Mass of hydrogen}/{Molecular mass of hydrogen} = 1/32

Hence, mass of hydrogen = 1/32 x molecular mass

= 1/32 x 2 x 10^-3 kg = 6.25 x 10^-5 kg

(a) The moon has small gravitational force and hence the escape velocity is small. As the moon is in the proximity of the Earth as seen from the Sun, the moon has the same amount of heat per unit area as that of the Earth. The air molecules have large range of speeds. Even though the rms speed of the air molecules is smaller than the escape velocity on the moon, a significant number molecules have speed greater than escape velocity and they escape. Now rest of the molecules arrange the speed distribution for the equilibrium temperature. Again a significant number of molecules escape as their speeds exceed escape speed. Hence, over a long time the moon has lost most of its atmosphere.

At 300 K,

\(V_{rms}=\sqrt\frac{3kT}{m}\) 

\(=\sqrt\frac{3\times1.38\times10^{-23}\times300}{7.3\times10^{-26}}\) 

= 1.7 km/s

V_esc for moon = 4.6 km/s.

(b) As the molecules move higher their potential energy increases and hence kinetic energy decreases and hence temperature reduces. At greater height more volume is available and gas expands and hence some cooling takes place.

The average K.E will be the same as conditions of temperature and pressure are the same 

v_rms α (1/√m)

m_A >m_B m_C 

v_C >v_B >v_A

When air is pumped, more molecules are pumped in. Boyle’s law is stated for situation where number of molecules remain constant.

(b) will not be valid since the ions would experience forces other than due to collisions with the walls.

(d) will not be valid because isotropy is lost.

We have 0.25 × 6 × 10²³ molecules, each of volume 10^-30m³.

Molecular volume = 2.56 x 10^-7m³ 

Supposing Ideal gas law is valid.

Final volume = (V_in/100) = ((3)³ x 10^-6)/100 ≈ 2.7 x 10^-7m³

which is about the molecular volume. Hence, intermolecular forces cannot be neglected. Therfore the ideal gas situation does not hold.

(c) only the translational part of the energy because during collisions with the wall pressure relates to change in linear momentum.

(b) Remains the same because motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls.

Comment for discussion: This brings in concepts of relative motion and that when collision takes place, it is the relative velocity which changes.

Brownian motion. In this motion any particular molecule will follow a zig-zag path due to the collisions with the other molecules or with the walls of the container.

The r.m.s. speed of the molecules of a gas is defined as the square root of the mean of the squared velocities of the molecules of gas. No, r.m.s. speed is different from the average speed.

\(c=v_{rms}=\sqrt\frac{v_1^2+v_2^2+v_3^2}{3}\) 

and \(v_{avg}=\frac{v_1+v_2+v_3}{3}\)

The average K.E. of molecule of a gas molecules depends only on the absolute temperature of the gas and is directly proportional to it.

According to kinetic theory of gases, the P exerted by a gas density ρ and r.m.s. velocity v is given by

P =  \(\frac{1}{3}\)ρv²

Mass per unit volume of gas = Density (ρ)

Average K.E. of translation per unit of volume of the gas.

\(E=\frac{1}{2}\)ρv²

^{\(\frac{P}{E}=\frac{\frac{1}{3}ρv^2}{\frac{1}{2}ρv^2}=\frac{2}{3}\)}

P= \(\frac{2}{3}E\)

= \(\frac{2}{3}\)× Average K.E. per unit volume.

Volume occupied by 1 gram mole of gas at NTP = 22400cc

∴ Number of molecules in 1cc of hydrogen

\(=\frac{6.023\times10^{23}}{22400}\)

= 2.688 × 10¹⁹

As each diatomic molecule has 5 degrees of freedom, hydrogen being diatomic also has 5 degrees of freedom.

∴ Total no. of degrees of freedom 

= 5 × 2.688 × 10¹⁹

= 1.344 × 10²⁰