The d for water vapour is same as that of air. The number density is inversely proportional to absolute temperature.
\(\text { So } n=2.7 \times 10^{25} \times \frac{273}{373}=2 \times 10^{25} \mathrm{~m}^{-3}\)
Hence, mean free path l = 4 x 10^–7m

Using the gas law PV = µRT, you can easily show that 1 mol of any (ideal) gas at standard temperature (273 K) and pressure (1 atm = 1.01 x 10⁵ Pa) occupies a volume of 22.4 litres. This universal volume is called molar volume. Thus the cylinder in this example contains 2 mol of helium. Further, since helium is monatomic, its predicted (and observed) molar specific heat at constant volume, C_v = (3/2) R, and molar specific heat at constant pressure, C_p = (3/2) R + R = (5/2) R. Since the volume of the cylinder is fixed, the heat required is determined by C_v . Therefore,
Heat required = no. of moles × molar specific heat × rise in temperature
= 2 x 1.5 R x 15.0 = 45 R
= 45 x 8.31 = 374 J.

(a) Let the speed of the ball be u relative to the wicket behind the bat. If the bat is moving towards the ball with a speed V relative to the wicket, then the relative speed of the ball to bat is V + u towards the bat. When the ball rebounds (after hitting the massive bat) its speed, relative to bat, is V + u moving away from the bat. So relative to the wicket the speed of the rebounding ball is V + (V + u) = 2V + u, moving away from the wicket. So the ball speeds up after the collision with the bat. The rebound speed will be less than u if the bat is not massive. For a molecule this would imply an increase in temperature.
You should be able to answer (b) (c) and (d) based on the answer to (a). (Hint: Note the correspondence, piston → bat, cylinder → wicket, molecule → ball.)

Partial pressure of a gas in a mixture is the pressure it would have for the same volume and temperature if it alone occupied the vessel. (The total pressure of a mixture of non-reactive gases is the sum of partial pressures due to its constituent gases.) Each gas (assumed ideal) obeys the gas law. Since V and T are common to the two gases, we have P₁V = µ₁ RT and P₂V = µ2 RT, i.e. (P₁ /P₂ ) = (µ₁ / µ₂ ). Here 1 and 2 refer to neon and oxygen respectively. Since (P₁ /P₂ ) = (3/2) (given), (µ₁ / µ₂ ) = 3/2.
(i) By definition µ₁ = (N₁ /NA ) and µ₂ = (N₂ /NA ) where N₁ and N₂ are the number of molecules of 1 and 2, and NA is the Avogadro’s number. Therefore, (N₁ /N₂ ) = (µ₁ / µ₂ ) = 3/2.
(ii) We can also write µ₁ = (m₁ /M₁ ) and µ₂ = (m₂ /M₂ ) where m1 and m2 are the masses of 1 and 2; and M₁ and M₂ are their molecular masses. (Both m1 and M₁ ; as well as m₂ and M₂ should be expressed in the same units). If ρ₁ and ρ2 are the mass densities of 1 and 2 respectively, we have \( \frac{\rho_1}{\rho_2}=\frac{m_1 / V}{m_2 / V}=\frac{m_1}{m_2}=\frac{\mu_1}{\mu_2} \times\left(\frac{M_1}{M_2}\right) \)
\(=\frac{3}{2} \times \frac{20.2}{32.0}=0.947 \)

(i) Given, T = 27 \(^0\)C
= ( 273.15 + 27 )
= 300.15K
Average thermal energy , E =  \(\frac{3}{2}\)k_BT
( where, kB = Boltzman constant
= 1.38\(\times\)10^-23 JK^-1 )
E = \(\frac{3}{2}\)\(\times\)1.38\(\times\)10^-23\(\times\)300.15
= 6.21 \(\times\)10^-21 J
(ii) At the temperatures , T = 107 K
Average thermal energy , E = \( \frac{3}{2}\)k_BT
= \( \frac{3}{2}\)\(\times\)1.38\(\times\)10^-23\(\times\)6000
= 1.241\(\times\)10^-19 J
(iii) At temperature , T = 10⁷K
Average thermal energy,
E = \(\frac{3}{2}\)k_BT
= \(\frac{3}{2}\)\(\times\)1.38\(\times\)10^-23\(\times\)10⁷
= 2.07\(\times\)10^-16 J

A given mass of water in vapour state has 1.67×10³ times the volume of the same mass of water in liquid state : For a given mass of water molecules, the density is less if volume is large. So the volume of the vapour is 1000/0.6 = 1/(6 × 10 ^-4 ) times larger. If densities of bulk water and water molecules are same, then the fraction of molecular volume to the total volume in liquid state is 1. As volume in vapour state has increased, the fractional volume is less by the same amount, i.e. 6×10^-4 .
This is also the increase in the amount of volume available for each molecule of water. When volume increases by 103 times the radius increases by V1/3 or 10 times, i.e., 10 × 2 Å = 20 Å. So the average distance is 2 × 20 = 40 Å.

In the liquid (or solid) phase, the molecules of water are quite closely packed. The density of water molecule may therefore, be regarded as roughly equal to the density of bulk water = 1000 kg m^–3. To estimate the volume of a water molecule, we need to know the mass of a single water molecule. We know that 1 mole of water has a mass approximately equal to
(2 + 16)g = 18 g = 0.018 kg.
Since 1 mole contains about 6 × 10²³ molecules (Avogadro’s number), the mass of a molecule of water is (0.018)/(6 × 10²³) kg = 3 × 10^–26 kg.
Therefore, a rough estimate of the volume of a water molecule is as follows :
Volume of a water molecule = (3 × 10^–26 kg)/ (1000 kg m^–3)
= 3 × 10^–29 m³ = (4/3) π (Radius)³
Hence, Radius ≈ 2 ×10^-10 m = 2 Å

For a given mass of water molecules, the density is less if volume is large. So the volume of the vapour is 1000/0.6 = 1/(6 × 10 ^-4 ) times larger. If densities of bulk water and water molecules are same, then the fraction of molecular volume to the total volume in liquid state is 1. As volume in vapour state has increased, the fractional volume is less by the same amount, i.e. 6×10^-4 .

At a fixed temperature the average energy = ½ m 2> is constant. So smaller the mass of the molecule, faster will be the speed. The ratio of speeds is inversely proportional to the square root of the ratio of the masses. The masses are 349 and 352 units. So
v₃₄₉ / v₃₅₂ = ( 352/ 349)^1/2 = 1.0044
Hence difference = \(\frac{\Delta V}{V}=0.44 \%\)
[ ²³⁵Uis the isotope needed for nuclear fission. To separate it from the more abundant isotope ²³⁸U, the mixture is surrounded by a porous cylinder. The porous cylinder must be thick and narrow, so that the molecule wanders through individually, colliding with the walls of the long pore. The faster molecule will leak out more than the slower one and so there is more of the lighter molecule (enrichment) outside the porous cylinder The method is not very efficient and has to be repeated several times for sufficient enrichment.

Volume of oxygen, V₁ = 30 litres = 30 x 10^–3 m³
Gauge pressure, P₁ = 15 atm = 15 x 1.013 × 10⁵ Pa
Temperature, T₁ = 27°C = 300 K
Universal gas constant, R = 8.314 J mole^–1 K^–1
Let the initial number of moles of oxygen gas in the cylinder be n₁.
The gas equation is given as:
\(P_{1} V_{1}=\eta_{1} R T_{1}\)
\(\therefore n_{1}=P_{1} \frac{V_{1}}{R} T_{1}\)
\(=\frac{15.195 \times 10^{5} \times 30 \times 10^{-3}}{(8.314) \times 300}=18.276\) 
Where,
m₁ = Initial mass of oxygen
M = Molecular mass of oxygen = 32 g
∴ m₁ = n₁M = 18.276 x 32 = 584.84 g
After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.
Volume, V₂ = 30 litres = 30 x 10^–3 m³
Gauge pressure, P₂ = 11 atm = 11 x 1.013 x 10⁵ Pa
Temperature, T₂ = 17°C = 290 K
Let n₂ be the number of moles of oxygen left in the cylinder.
The gas equation is given as:
P₂V₂ = n₂RT₂
\(\therefore n_{2}=\frac{P_{2} V_{2}}{R T_{2}}\)
\(=\frac{11.143 \times 10^{5} \times 30 \times 10^{-3}}{8.314} \times 290=13.86\)
\(\text { But } n_{2}=\frac{m_{2}}{M}\)
Where,
m₂ is the mass of oxygen remaining in the cylinder
∴ m₂ = n₂M = 13.86 x 32 = 453.1 g
The mass of oxygen taken out of the cylinder is given by the relation:
Initial mass of oxygen in the cylinder – Final mass of oxygen in the cylinder
= m₁ – m₂
= 584.84 g – 453.1 g
= 131.74 g
= 0.131 kg
Therefore, 0.131 kg of oxygen is taken out of the cylinder.

Here volume of room V = 25.0 m³ , temperature, T = 27⁰C = 300 K and
Pressure, P = 1 atm  = 1.01 x 10⁵ pa
According to gas equation,
PV = \(\mu \)RT = \(\mu \)N_A.K_B T
Hence, total number of air molecules in the volume of given gas,
N=\(\mu \). N_A = \(\frac { PV }{ { k }_{ B }T } \)
\(\therefore N=\frac { 1.01\times { 10 }^{ 5 }\times 25.0 }{ (1.38\times { 10 }^{ -23 })\times 300 } =6.1\times { 10 }^{ 26 }\) 

(a) The dotted plot corresponds to 'ideal' gas behaviour as it is parallel to P-axis and it tells that value of PV /T remains same even when P is changed
(b) The upper position of PV /T shows that its value is lesser for T₁, thus T₁> T_2. This is because the curve at T₁ is more close to dotted plot than the curve at T₂. Since the behaviour of a real gas approaches the perfect gas behaviour, as the tempera lure is increased
(c) Where the two curves meet, the value of PV / T on y-axis is equal to \(\mu \)R Since ideal gas
equation for \(\mu \) moles is PV = \(\mu \)RT
\(where,\ \mu =\frac { 1.00\times 10^{ -3 }kg }{ 32\times 10^{ -3 }kg } =\frac { 1 }{ 32 } \)
\(\therefore \ value\ of\frac { PV }{ T } \mu R=\frac { 1 }{ 32 } \times 8.31\quad { JK }^{ -1 }\)
(d) If we obtained similar plots for 1.00 x 10^-3 kg of hydrogen, we will not get the same value of \(\frac { PV }{ T } \)  at the point, where the curves meet on the y-axis. This is because molecular mass of hydrogen is different from that of oxygen.
For the same value of \(\frac { PV }{ T } \) mass of hydrogen required is obtained from
\(\frac { PV }{ T } nR=\frac { m }{ 2.02 } \times 8.31=0.26\)
\(\\ m=\frac { 2.02\times 0.26 }{ 8.31 } gram=6.32\times { 10 }^{ -2 }gram\) 

Let n the number of molecular per unit vollume of the gas.pV = NkT, where N = number of molecules in volume of the gas
\(n=\frac { N }{ V } =\frac { P }{ kT } \)
Here,
\(p=2\times 1.01\times { 10 }^{ 5 }N/m^{ 2 },k=1.38\times 10^{ -23 }j/K\)
\(\\ T=273\times 17=290K\)
\(\\ \ \therefore \ n=\frac { 2\times 1.01\times { 10 }^{ 5 } }{ (1.38\times { 10 }^{ -23 })\times (290) } \)
\(\\ =5.05\times { 10 }^{ 25 } molecule/m^{ 3 }\)
 Mean free path
\(\lambda =\frac { 1 }{ 4\pi \sqrt { 2 } { r }^{ 2 }n } \)
Here,
\(r=1.0\mathring { A=1.0\times { 10 }^{ -19 }m }\)  
\( \\ \therefore \lambda =\frac { 1 }{ 4\pi \sqrt { 2 } (1.0\times { 10 }^{ -10 })\times 5.05\times { 10 }^{ 25 } } \)
\( \lambda =1.11\times { 10 }^{ -7 }m\)
Now , \({ v }_{ rms }=\sqrt { \frac { 3RT }{ M } } =\sqrt { \frac { 3\times 8.31\times 290 }{ 28\times { 10 }^{ -3 } } } =508.14m/s\)
\(\therefore \)  collision frequency,
 \(f=\frac { { v }_{ rms } }{ \lambda } =\frac { 508.14 }{ 1.11\times { 10 }^{ -7 } } \)
 \(=4.58\times 10^{ 9 }\)collissions/s     

\({ V }_{ 1 }=1\ { cm }^{ 3 }=1\times { 10 }^{ -6 }{ m }^{ 3 },\)
\(\\ { T }_{ 1 }={ 12 }^{ \circ }=273+35=308K\)
\(\\ { P }_{ 1 }=1\ atm\ +{ h }_{ 1 }pg\)
\(=1.01\times { 10 }^{ 5 }+40\times { 10 }^{ 3 }\times 9.8\ =493000\ Pa\)
Initially, when air bubble is at 40 m deep, Then,          
Firstly, when air bubble reaches at the surface, then
 \( { T }_{ 2 }={ 35 }^{ \circ }C=273+35\ =308K\)
\({ P }_{ 2 }=1\ atm=1.01\times { 10 }^{ 5 }\ Pa,{ V }_{ 2 }=?\)
\(As\ we\ know,\ \frac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\frac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \Rightarrow { V }_{ 2 }=\frac { { P }_{ 1 }{ V }_{ 1 }{ T }_{ 2 } }{ { T }_{ 1 }{ P }_{ 2 } } \)
Final volume
\(\\ { V }_{ 2 }=\frac { \left( 493000 \right) \times \left( 1\times { 10 }^{ -6 } \right) \times 308 }{ 285\times 1.01\times { 10 }^{ 5 } } \)
\(\\ =5.275\times { 10 }^{ -6 }{ m }^{ 2 }\) 

let C and C' be the rms velocity of argon and a helium gas atoms at temperature TK and T'K respectively.
Here, M = 39.9 M '= 4.0, T = ?.
T' = -20 + 273 = 253 K
Now,  v = \(\sqrt { \frac { 3RT }{ M } }\)
= \(\sqrt { \frac { 3RT }{ 39.9 } }\) and v' = \(\sqrt { \frac { 3R{ T }^{ ' } }{ { M }^{ ' } } }\)
= \(\sqrt { \frac { 3R\times 253 }{ 4 } }\)
Since, v = v',
\(\therefore\) \(\sqrt { \frac { 3R{ T } }{ 39.9 } }\) =\(\sqrt { \frac { 3R\times 253 }{ 4 } }\)
or    T = \(\frac { 39.9\times 253 }{ 4 } \)
= 2523.7 K

As three vessels are identical i.e., they have same volume now at constant pressure, temperature and volume the three vessels will contain equal number of molecules (by Avogadro’s law) and is equal to Avogadro's number, NA = 6.0²³ x 10²³
\(\because { V }_{ rms }=\sqrt { \frac { 3{ k }_{ B }T }{ m } } \Rightarrow { V }_{ rms }\propto \frac { 1 }{ \sqrt { m } }\)
where, m is mass of single gas molecule as neon has the smallest mass, so rms speed will be greatest in case of neon.

The important point to remember is that the average kinetic energy (per molecule) of any (ideal) gas (be it monatomic like argon, diatomic like chlorine or polyatomic) is always equal to (3/2) k_B T. It depends only on temperature, and is independent of the nature of the gas.
(i) Since argon and chlorine both have the same temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is 1:1.
(ii) Now ½ m v_rms² = average kinetic energy per molecule = (3/2) ) k_B T where m is the mass of a molecule of the gas. Therefore,
\( \frac{(v^2_rms)_{Ar}}{(v_rms)cl_2}\) = \( \frac{(m)cl}{(m)_{Ar}}\) =\(\frac{M_{cl}}{M_{Ar}}\) = \(\frac{70.9}{39.9}\) = 1.777
where M denotes the molecular mass of the gas. (For argon, a molecule is just an atom of argon.) Taking square root of both sides,
\(\frac{(v_{rms})_{Ar}}{(v_{rms})_cl_2}\) = 1.333
You should note that the composition of the mixture by mass is quite irrelevant to the above calculation. Any other proportion by mass of argon and chlorine would give the same answers to (i) and (ii), provided the temperature remains unaltered.

Given, d =  \(3\overset { o }{ A }.\)
r = \(\frac{d}{2}\)
= 1.5 \(\times\) 10-10 m
= 1.5 \(\times\) 10-8 cm
Molecular volume, V =\( \frac{4}{3}\) \(\pi\) r³N
= \(\frac{4}{3}\) \(\times\)3.14\(\times\)(1.5\(\times\)10^-8)³\(\times\)6.023\(\times\)10²³
= 8.52 cc
Actual volume occupied by 1 mole of oxygen at STP,
V' = 22400 cc
\(\therefore\) \(  \frac{V}{V'} \) = \(\frac{8.52}{22400}\)
= 3.8 \(\times\)10^-4

As for one mole of ideal gas,  pV =\(\mu\)RT
pV = RT
V = \(\frac{RT}{p}\)
Putting , R = 8.31 Jmol^-1K^-1, T = 273 K
p = 1 atm = 1.013 \(\times \)10⁵ N -m^-2
v =  \(\frac{8.31\times273}{1.013\times10^5}\)
= 0.0224 m³
= 22.4 L  [ \(\therefore\) 1 m³ = 10³L ]