From kinetic theory of gases, the pressure P exerted by an ideal gas of density p and r.m.s. velocity of its gas molecules C is given by
\(p=\frac { 1 }{ 3 } \rho C^{ 2 }\)
Mass of unit volume of the gas = 1 x P = P
Mean kinetic energy of translation per unit volume of the gas is
\(E=\frac { 1 }{ 2 } \rho C^{ 2 }\)
\(\\ \therefore \frac { P }{ E } =\frac { (1/3)\rho C^{ 2 } }{ (1/2)\rho C^{ 2 } } =\frac { 2 }{ 3 } \)
\(\\ or\ P=\frac { 2 }{ 3 } E\)
The pressure exerted by an ideal gas is numerically equal to two third of the mean kinetic energy of translation per unit volume of the gas. " Average Kinetic Energy per Molecule of the Gas
Consider one gram mole of an ideal gas occupying a volume V at temperature T. Let m be the mass of each molecule of the gas. Then
M = m x N_A
where N_A is Avogadro's number.
If C is the r.m.s. velocity of the gas molecules, then pressure P exerted by ideal gas is
\(P=\frac { 1 }{ 3 } { PC }^{ 2 }=\frac { 1 }{ 3 } \frac { M }{ V } { C }^{ 2 }\)
\(\\ or\ PV=\frac { 1 }{ 3 } M{ C }^{ 2 }\)
From perfect gas equation, PV = RT, where R is a universal gas constant for one gram mole of the gas
\(\therefore \) \(\frac { 1 }{ 3 } { MC }^{ 2 }=RT\quad OR\quad \frac { 1 }{ 3 } { MC }^{ 2 }=\frac { 3 }{ 2 } RT\)
\(\therefore \) Average kinetic energy of translation of one mole of the gas
\(\frac { 1 }{ 3 } { MC }^{ 2 }=\frac { 3 }{ 2 } RT\)
\(or\frac { 1 }{ 2 } mN_{ A }{ C }^{ 2 }=\frac { 3 }{ 2 } RT\)
\(\\ or\ \frac { 1 }{ 2 } mC^{ 2 }=\frac { 3 }{ 2 } \left( \frac { R }{ N_{ A } } \right) T=\frac { 3 }{ 2 } k_{ B }T\)
where k₈ is called Boltzmann constant.
\(\therefore \) Average K.E. of translation per molecule of gas
\(\frac { 1 }{ 2 } mC^{ 2 }=\frac { 3 }{ 2 } k_{ B }T\) 

When the tube is held horizontally, the mercury thread of length 76 cm traps a length of air = 15 cm. A length of 9 Cm of the tube will be left at the open end. The pressure of air enclosed in
tube will be atmospheric pressure. Let area of cross-section of the tube be 1sq. cm.
\(\therefore \) P₁ = 76 Cm  and V₁ = 15 Cm³
When the tube is held vertical1y, 15 em air gets another 9 cm of air (filled in the right handside in the horizontal position) and let h cm of mercury flows out to balance the atmospheric pressure. Then the heights of air column and mercury column are (24 + h) cm and (76 - h) cm respectively.
The pressure of air  = 76 - (76 - h) -h cm of mercury.
\(\therefore \)  V₂ = (24 + h) CM³  and P₂ = h cm
If we assume that temperature remains constant, then
P₁V₁ =P₂ V₂ or 76 x 15 = h x (24 + h) or h² + 24h - 1140 = 0
\(or \ h=\frac { -24\pm \sqrt { \left( { 24 } \right) ^{ 2 }+4\times 1140 } }{ 2 } =23.8\ cm\ or\ -47.8cm\)
Since h cannot be negative(because more mercury cannot flow into the tube), therefore h = 23.8 cm thus in the vertical position of the tube, 23.8 cm of mercury flows out.

Let
CV´= molar heat capacity of the first gas,
CV´´ = molar heat capacity of the second gas,
CV = molar heat capacity of the mixture,
and similar symbols for other quantities. Then,
γ= C′p/C′V=1.67
and
C′P=C′V+R
This gives
C′V=3/2R and C′P=5/2R
Similarly, γ = 1.4 gives C′′V=5/2R
and C′′P=7/2R
Suppose the temperature of the mixture is increased by dT. The increase in the internal energy of the first gas = n₁C′VdT.
The increase in internal energy of the second gas = n₂C′′VdT.
Thus, \(\left(n_1+n_2\right) C_V d T=n_1 C_V^{\prime} d T+n_2 C V " d T \)
or, \(C_V=\frac{n_1 C_v^{\prime}+n_2 C_v^{\prime \prime}}{n_1+n_2}\)
\( C_P=C_V+R=\frac{n_1 C_v^{\prime}+n_2 C_v^{\prime \prime}}{n_1+n_2}+R \)
\( =\frac{n_1\left(C_v^{\prime}+R\right)+n_2\left(C_v^{\prime \prime}+R\right)}{n_1+n_2} \)
\(=\frac{n_1 C_p^{\prime}+n_2 C_p^{\prime \prime}}{n_1+n_2} \ldots(2)\)
From (1) and (2),
\( Y=\frac{C_p}{C_V}=\frac{n_1 C_p^{\prime}+n_2 C_p^{\prime \prime}}{n_1 C_V^{\prime}+n_2 C_V^{\prime \prime}} \)
\( \frac{4 \times \frac{5}{2} R+2 \times \frac{7}{2} R}{4 \times \frac{3}{2} R+2 \times \frac{5}{2} R}=1.54 \)

(i) Average speed
\(=\frac { { n }_{ 1 }{ v }_{ 1 }+{ n }_{ 2 }{ v }_{ 2 }+{ n }_{ 3 }{ v }_{ 3 }+{ n }_{ 4 }{ v }_{ 4 }+{ n }_{ 5 }{ v }_{ 5 } }{ n_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+{ n }_{ 4 }+{ n }_{ 5 } }\) 
\( \\ =\frac { 2\times 1+4\times 2+8\times 3+6\times 4+3\times 5 }{ 2+4+8+6+3 }\)
\( \\ =3.17\ m/s\)
(ii) Root mean square speed
\(=\sqrt { \frac { { n }_{ 1 }{ v }_{ 1 }^{ 2 }+{ n }_{ 2 }{ v }_{ 2 }^{ 2 }+{ n }_{ 3 }{ v }_{ 3 }^{ 2 }+{ n }_{ 4 }{ v }_{ 4 }^{ 2 }+n_{ 5 }{ v }_{ 5 }^{ 2 } }{ { n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+{ n }_{ 4 }+{ n }_{ 5 } } } \)
\(\\ =\sqrt { \frac { 2\times { 1 }^{ 2 }+4\times { 2 }^{ 2 }+8\times { 3 }^{ 2 }+6\times 4^{ 2 }+3\times { 5 }^{ 2 } }{ 2+4+8+6+3 } }\)
\(=3.36\ m/s\)
(iii)The most probable speed is that speed which is possessed by maximum of molecules.
 \(Most\ protable\ speed\ ({ V }_{ mp })=\sqrt { \frac { 2{ K }_{ B }T }{ m } } =\sqrt { \frac { 3{ K }_{ B }T }{ 3 } \times 2/3 } \)
\(\\ { v }_{ mp }=\sqrt { \frac { 2 }{ 3 } } \times \sqrt { \frac { 3{ K }_{ B }T }{ 3 } } =\sqrt { \frac { 2 }{ 3 } { V }_{ rms } } =\sqrt { \frac { 2 }{ 3 } } \times 3.36\ m/s\)
\(=0.816\times 3.36\ m/s\ =2.74\ m/s\)

We know, that density of an element
\(\rho =\frac { Mass }{ Volume } =\frac { Mass\ of\ 1\ mole }{ Total\ volume\ of\ molecules\ in\ 1\ mole\ when\ closely\ packed } \)
\( \rho =\frac { M(in\ grams) }{ \left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) { N }_{ A } } =\frac { 3M\times { 10 }^{ -3 }kg }{ 4\pi { r }^{ 3 }.{ N }_{ A } }\)
\( \\ \Rightarrow r=\left[ \frac { 3M\times { 10 }^{ -3 } }{ 4\pi { N }_{ A }.\rho } \right] ^{ \frac { 1 }{ 3 } },\)
\(\\ where,{ N }_{ A }=Avogardo's\ number\simeq 6\times { 10 }^{ 23 }\)
(i) \(For\quad carbon\ (diamond),\)
\(\\ M=12.01,\rho =2.22\times { 10 }^{ 3 }kg/{ m }^{ 3 }\)
\(\\ \therefore Radius\ of\ carbon\ atom\)
\(\\ r=\left[ \frac { 3\times 12.01\times { 10 }^{ -3 } }{ 4\times 3.14\times 6\times \times { 10 }^{ 23 }\times 2.22\times { 10 }^{ 3 } } \right] ^{ \frac { 1 }{ 3 } }\)
\(\\ \ =1.29\times { 10 }^{ -10 }m=1.29\overset { 0 }{ A } .\) 
(ii) For gold, M = 197.00,p = 19.32 x 10 ³ kg/m³
\(\therefore\) Radius of gold atom
\(r=\left[\frac{3 \times 197 \times 10^{-3}}{4 \times 3.14 \times 6 \times 10^{23} \times 19.32 \times 10^{3}}\right]^{\frac{1}{3}}\)
= 1.59 x 10^{- 10} m = 1.59 \(\overset { 0 }{ A } \)
(iii) For nitrogen (liquid)
M = 14.01, P = 1.00 x 10³ kg / m³
\(\therefore\) Radius of nitrogen atom
\(r=\left[\frac{3 \times 14.01 \times 10^{-3}}{4 \times 3.14 \times 6 \times 10^{23} \times 1.0 \times 10^{3}}\right]^{ \frac { 1 }{ 3 } }\)
= 1.77 x 10^{- 7} m = 1.77 \(\overset { 0 }{ A } \) 
(iv) For lithium, M = 6.94, P = 0.53 c 10³ kg/ m³
\(\therefore\) Radius of lithium atom
\(r=\left[\frac{3 \times 6.94 \times 10^{-3}}{4 \times 3.14 \times 6 \times 10^{23} \times 0.53 \times 10^{3}}\right]^{ \frac { 1 }{ 3 } }\)
= 1.73 x 10^{- 10} m = 1.73 \(\overset { 0 }{ A } \)
(v) For flourine (liquid),
M = 19.00 and p = 1.14 x 10³ kg / m³
\(\therefore\) Radius of fluorine atom
\(r=\left[\frac{3 \times 19.0 \times 10^{-3}}{4 \times 3.14 \times 6 \times 10^{23} \times 1.14 \times 10^{3}}\right]^{\frac{1}{3}}\)
= 1.88 x 10^{- 10} m= 1.88 \(\overset { 0 }{ A } \)

Given, volume of the box, V = 1.00 m ³
Area, a = 0.010 mm² = 10 ^-8m²

Temperature outside = Temperature inside
Initial pressure inside the box = 1.50 atm.
Final pressure inside the box = 0.10 atm
Assuming,
v_ix = Speed of nitrogen molecule inside the box along x-direction
n₁= Number of molecules per unit volume in a time interval of  \(\triangle\)t, all the particles at a distance (v_ix\(\triangle\) t) will collide the hole and the wall, the particle colliding along the hole will escape out reducing the pressure in the box.
Let area of the wall is A, number of particles colliding in time,  \(\triangle\)t
\(=\frac{1}{2} n_{1}\left(v_{i x} \Delta t\right) A\) 
\(\frac{1}{2}\) is the factor because all the particles along x- direction are behaving randomly. Hence, half of these are colliding against the walls on either side.
Inside the box \(v_{i x}^{2}+v_{i y}^{2}+v_{i z}^{2}=v_{\mathrm{rms}}^{2}\)
\(\Rightarrow \quad v_{i x}^{2}=\frac{v_{\mathrm{rms}}^{2}}{3} \quad\left[\because v_{i x}=v_{i y}=v_{i z}\right]\)
If particles collide along hole, they move out. Similarly, outer particles colliding along hole will move in.
If a = area of hole 
Then, net particle flow in time, 
\(\Delta t=\frac{1}{2}\left(n_{1}-n_{2}\right) \frac{k_{B} T}{m} \Delta t a \quad\left[\because v_{\mathrm{rms}}=\sqrt{\frac{3 k_{B} T}{m}}\right]\)
[Temperature inside and outside the box are equal]
Let n = number of density of nitrogen
\(n=\frac{\mu N_{A}}{V}=\frac{p N_{A}}{R T} \quad\left[\because \frac{\mu}{V}=\frac{p}{R T}\right]\)
Where, N_A = Avogadro's number
If after time \(\triangle\)t , pressure inside changes from P to p'1
\(\therefore \quad n_{1}^{\prime}=\frac{p_{1}^{\prime} N_{A}}{R T}\) 
Now, number of molecules gone out = \(n_{1} V-n_{1}^{\prime} V\)
\(=\frac{1}{2}\left(n_{1}-n_{2}\right) \sqrt{\frac{k_{B} T}{m}} \Delta t a\)
\(\therefore \frac{p_{1} N_{A}}{R T} V-\frac{p_{1}^{\prime} N_{A}}{R T} V=\frac{1}{2}\left(p_{1}-p_{2}\right) \frac{N_{A}}{R T} \sqrt{\frac{k_{B} T}{m}} \Delta t a\)
\(\Rightarrow \quad \Delta t=2\left(\frac{p_{1}-p_{1}^{\prime}}{p_{1}-p_{2}}\right) \frac{V}{a} \sqrt{\frac{m}{k_{B} T}}\)
Putting the values from the data given,
\(\Delta t=2\left(\frac{1.5-1.4}{1.5-1.0}\right) \frac{1 \times 1.00}{0.01 \times 10^{-6}} \sqrt{\frac{46.7 \times 10^{-27}}{1.38 \times 10^{-23} \times 300}}\)
\(=\frac{2}{5} \times 3.358 \times 10^{5}=\frac{6.717}{5} \times 10^{5}\) 
\(\tau\) = 1.34 x 10⁵ s.