Given V₁ = 4 L, V₂ =?, P₁ = 100 N/m²  , P₂ =150N/m², \(\Delta V=?\)
Using Boyle's law for constant temperature, we have
\({ p }_{ 1 }{ V }_{ 1 }={ p }_{ 2 }{ V }_{ 2 }\)
\(\\ \Rightarrow { V }_{ 2 }=\frac { { p }_{ 1 }{ V }_{ 1 } }{ { p }_{ 2 } } =\frac { 100\times 4 }{ 150 } =2.667\ L\)
\(\therefore \) Change in volume, \(\triangle V={ V }_{ 1 }-{ V }_{ 2 }\ =4-2.667=1.33\ L\)

As two non-reactive gases are mixed in a container, So the value of V and T will be same for both with partial pressures, P₁ and P₂.
For gas I   \({ p }_{ 1 }V={ \mu }_{ 1 }RT\)
For gas II  \({ p }_{ 2 }V={ \mu }_{ 2 }RT\)
Dividing Eq. (i) byEq. (ii), we get
\(\frac { { p }_{ 1 } }{ { p }_{ 2 } } =\frac { { \mu }_{ 1 } }{ { \mu }_{ 2 } }\)
\( \\ { \mu }_{ 1 }={ N }_{ 1 }/{ N }_{ A }\)
\(\\ { \mu }_{ 2 }={ N }_{ 2 }/{ N }_{ A }\)
Dividing Eq. (iv) by Eq. (v)
\(\frac { { \mu }_{ 1 } }{ { \mu }_{ 2 } } =\frac { { N }_{ 1 } }{ { N }_{ 2 } } \)
From Eqs. (iii) and (vi) we get
\(\frac { { N }_{ 1 } }{ { N }_{ 2 } } =\frac { { \mu }_{ 1 } }{ { \mu }_{ 2 } } =\frac { { p }_{ 1 } }{ { p }_{ 2 } } \Rightarrow \frac { { N }_{ 1 } }{ { N }_{ 2 } } =\frac { { p }_{ 1 } }{ { p }_{ 2 } } \quad or\quad \frac { { N }_{ 1 } }{ { N }_{ 2 } } =\frac { 5 }{ 3 } \) 

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.

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
\(\frac { 1 }{ 2 } mC^{ 2 }=\frac { 3 }{ 2 } k_{ B }T\) 

We know that r.m.s. speed is given by
\(\\ { v }_{ rms }=\left[ \frac { 3PV }{ M } \right] ^{ 1/2 }=\left[ \frac { 3RT }{ M } \right] ^{ 1/2 }\)
If (v_rms)l be the r.m.s. speed of oxygen and (v_rms)be the r.m.s. of helium gas at temperature T₁ and T₂ respectively.
\({ (v }_{ rms })_{ 1 }=\left[ \frac { 3RT_{ 1 } }{ { M }_{ 1 } } \right] ^{ 1/2 }\quad and\quad { (v }_{ rms })_{ 2 }=\left[ \frac { 3RT_{ 2 } }{ { M }_{ 2 } } \right] ^{ 1/2 }\)
\(or\quad \frac { { (v }_{ rms })_{ 1 } }{ { (v }_{ rms })_{ 2 } } =\left[ \frac { { M }_{ 2 }{ T }_{ 1 } }{ { M }_{ 1 }{ T }_{ 1 } } \right] ^{ 1/2 }\)
here \({ (v }_{ rms })_{ 1 }={ (v }_{ rms })_{ 2 },{ M }_{ 1 }=32,{ M }_{ 2 }=4.0;\)
T₁ = ?,T₂ = -10 + 273 = 263K
\(\\ \therefore \ 1=\left[ \frac { 4\times { T }_{ 1 } }{ 32\times 263 } \right] ^{ 1/2 }\)
\(\\ or\quad { T }_{ 1 }=\frac { 32\times 263 }{ 4 } =2104K\)

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 = \(\frac { PV }{ { k }_{ B }T } \)
\(\therefore N=\frac { 1.01\times { 10 }^{ 5 }\times 25.0 }{ (1.38\times { 10 }^{ -23 })\times 300 } =6.1\times { 10 }^{ 26 }\) 

Given, \({ m }_{ 1 }=m,{ \quad v }_{ 1 }=v,\quad { V }_{ 1 }=V,\quad { P }_{ 1 }=?\)
 \({ m }_{ 2 }=\frac { m }{ 2 } ,\ { v }_{ 2 }=2v,\ { V }_{ 2 }={ V }_{ 1 }=V,\quad { p }_{ 2 }=?\)
\( { P }_{ 1 }=\frac { 1 }{ 3 } \frac { { m }_{ 1 } }{ { V }_{ 1 } } { v }_{ 1 }^{ 2 }\ and\ { p }_{ 2 }=\frac { 1 }{ 3 } \frac { { m }_{ 2 } }{ { V }_{ 2 } } { v }_{ 2 }^{ 2 }\)
\(\\ \therefore \frac { { P }_{ 1 } }{ { p }_{ 2 } } =\frac { { m }_{ 1 } }{ { m }_{ 2 } } \times \frac { { V }_{ 1 } }{ { V }_{ 2 } } \times \frac { { v }_{ 1 }^{ 2 } }{ { v }_{ 2 }^{ 2 } } \)
\(\\ =\frac { m }{ m/2 } \times \frac { V }{ V } \times \left( \frac { v }{ 2v } \right) ^{ 2 }\)
\(=2\times \frac { 1 }{ 4 } =\frac { 1 }{ 2 } \)

According to kinetic theory of gases,
\(pV=\frac { 1 }{ 3 } { Mv }_{ rms }^{ 2 }\)
At constant temperature, \(v_{ rms }^{ 2 }\) is constant. As v is also constant.
\(\therefore p\propto M\)
When the mass of the gas is increased by 50%, pressure also increased by 50%.
Resultant pressure = \(76+\frac { 50 }{ 100 } \times 76=114\quad cm\quad of\quad Hg\)

\( V=250 \mathrm{cc}=250 \times 10^{-6} \mathrm{~m}^3 \)
\( \mathrm{P}=10^{-3 \mathrm{~mm}}=10^{-3} \times 10^{-3} \mathrm{~m} \)
\( =\left(10^{-6} \times 13600 \times 10\right) \)
\(=136 \times 10^{-3} \text { Pascal } \)
\( \mathrm{T}=27^0 \mathrm{C}=300 \mathrm{k} \)
\( \mathrm{n}=\frac{\mathrm{PV}}{R T} \)
\( =\frac{136 \times 10^{-3} \times 250 \times 10^{-6}}{8.3 \times 300}=1.36 \times 10^{-8}\)
No. of molecules
\(=1.36 \times 10^{-8} \times 6 \times 10^{23}  \)
\(=8.17 \times 10^{15}\)

Initially in the oxygen cylinder, V₁=30 litre =\(30 \times 10^{-3} \mathrm{~m}^3\),
\(P_1=15 a t m=15 \times 1.01 \times 10^5 \mathrm{~Pa}, T_1=27+273 =300 \mathrm{~K}\)
If the cylinder contains n₁ mole of oxygen gas, then \(P_1 V_1=n_1 R T_1\)
or
\(n_1=\frac{P_1 V_1}{R T_1}=\frac{\left(15 \times 1.01 \times 10^5\right) \times\left(30 \times 10^{-3}\right)}{8.3 \times 300} =18.253\)
For oxygen moleculaer weight, M=32 g
Initial mass of oxygen in the cylinder cylinder,
\(m_1=n_1 M=18.253 \times 32=548.1 g\)
Finally in the oxygen gas in the cylinder, let n₂ moles of oxygen be left,
Here,
\( V_2=30 \times 10^{-3} \mathrm{~m}^3, P_2=11 \times 1.01 \times 10^5 \mathrm{~Pa}, T_2 =17+273=290 \mathrm{~K}\)
Now,
\( n_2=\frac{P_2 V_2}{R T_2}=\frac{\left(11 \times 1.01 \times 10^5\right) \times\left(30 \times 10^{-3}\right)}{8.3 \times 290} =13.847 \)
\( \therefore \text { Final mass of oxygen gas in the cylinder, } m_2=13.847 \times 32=453.1 \mathrm{~g} \)
\( \therefore \text { Mass of the oxygen gas withdrawn }=m_1-m_2=584.1-453.1=131.0 \mathrm{~g} . \)