According to kinetic theory of gases,
\(p=\frac { 1 }{ 3 } { P }^{ { C }^{ 2 } }=\frac { 1 }{ 3 } \frac { M }{ V } { C }^{ 2 }\)
\(=\frac { 1 }{ 3 } \frac { M }{ V } KT\)   
\( T\propto PV\)      \((\because { C }^{ 2 }=KT.\)When k is constant)
Now as T is directly proportional to the product of P and V. If P and V are constant, then T is also constant.

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 \( \frac{\rho_1}{\rho_2}=\frac{m_1 / V}{m_2 / V}=\frac{m_1}{m_2}=\frac{\mu_1}{\mu_2} \times\left(\frac{M_1}{M_2}\right) \)
\(=\frac{3}{2} \times \frac{20.2}{32.0}=0.947 \)

We know that
\({ C }^{ 2 }=3\frac { RT }{ Nm } =3\frac { kT }{ m } \)
Thus \({ C }_{ 1 }^{ 2 }=\frac { { 3kT }_{ 1 } }{ m } \)
and \({ C }_{ 2 }^{ 2 }=\frac { { 3kT }_{ 2 } }{ m } \)
\(\frac { { C }_{ 1 }^{ 2 } }{ { C }_{ 2 }^{ 2 } } =\frac { { T }_{ 1 } }{ { T }_{ 2 } } \)
Here C₂ = 2C₁,T - 273 + 100 = 373 K
T₂ = T₁ x \(\frac { { C }_{ 2 }^{ 2 } }{ { C }_{ 1 }^{ 2 } } \) = 373 x 4
= 1492 K = 1219⁰C

Given, \({ T }_{ 1 }=300k,{ T }_{ 2 }=?\)
\(\\ { V }_{ 1 }=V,{ V }_{ 2 }=\frac { 4 }{ 3 } V\)
According to Charles law, we get
\(\frac { { V }_{ 2 } }{ { V }_{ 1 } } =\frac { { T }_{ 2 } }{ { T }_{ 1 } } \)
\(\\ \Rightarrow { T }_{ 2 }={ T }_{ 1 }\frac { { V }_{ 2 } }{ { V }_{ 1 } } =300\times \frac { 4 }{ 3 } =400\ k\)

0.5

500 K

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 .

According to the law of atmospheres.
\({ n }_{ 2 }={ n }_{ 1 }\quad exp.\left[ -\frac { mg }{ { K }_{ B }T } \left( h_{ 2 }-{ h }_{ 1 } \right) \right] --\quad (i)\)
where, n₂ and n₁refer to number density of particles at heights h₂  and  h_1,  respectively.
If we consider the sedimentation equilibrium of suspended particles in a liquid, then in place of mg, we will have to take effective weight of the suspended particles.
Let, V = average volume of a suspended particle,
\(\rho \) =  density of suspended particle,  \({ \rho }^{ ' }\)= density of liquid, m = mass of one suspended particle, \({ m }^{ ' }\)= mass of equal volume of liquid displaced.
According to Archimedes' priciple, effective weight of one suspended particle
= Actual weight-weight of liquid displaced = mg-m^'g
\(=mg-V{ \rho }^{ ' }g=mg-\left( \frac { m }{ \rho } \right) { \rho }^{ ' }g=mg\left( 1-\frac { { \rho }^{ ' } }{ \rho } \right) \)
\(\\ Also,\ Boltzmann\ constant,{ K }_{ B }=\frac { R }{ { N }_{ A } } \)
where, R is gas constant and N_A is Avogardro's number.
\(putting,\ mg\left( 1-\frac { { \rho }^{ ' } }{ \rho } \right) in\ place\ of\ mg\ and\ value\ of\ { K }_{ B }\quad in\)
\( Eq.(i)\ we\ get\)
\( { n }_{ 2 }={ n }_{ 1 }exp\left[ -\frac { mg{ N }_{ A } }{ RT } \left( 1-\frac { { \rho }^{ ' } }{ \rho } \right) \left( { h }_{ 2 }-{ h }_{ 1 } \right) \right] ,\ which\ id\ required\ relation.\) 

Both gases have same value of average translational kinetic energy per mole because their temperatures are equal and \(\bar { { E } } =\frac { 3 }{ 2 } RT\)

(i) Low pressure
(ii) High temperature

Since  \(C\propto \sqrt { T } \)
Clearly, C will be doubled

The law of equipartition of energy for any dynamical system in thermal equilibrium, the total energy is distributed q = equally amongst all the degrees of freedom.
The energy associated with each molecule per degree of freedom is \(\frac { 1 }{ 2 } { k }_{ B }T\), where K_B is Boltzmann's constant and T is temperature of the system.

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.