In $$bcc$$  unit cell, the number of atoms = 2
Thus, volume of atoms in unit cell $$\left( v \right) = 2 \times \frac{4}{3}\pi {r^3}$$
For $$bcc$$  structure $$\left( r \right) = \frac{{\sqrt 3 }}{4}a$$
$$\eqalign{ & \left( V \right) = 2 \times \frac{4}{3}\pi {\left( {\frac{{\sqrt 3 }}{8}a} \right)^3} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{\sqrt 3 }}{8}\pi \,{a^3} \cr} $$
Volume of unit cell $$\left( V \right) = {a^3}$$
Percentage of volume occupied by unit cell
$$\eqalign{ & = \frac{{{\text{Volume of the atoms in unit cell}}}}{{{\text{Volume of unit cell}}}} \cr & = \frac{{\frac{{\sqrt 3 }}{8}\pi \,{a^3}}}{{{a^3}}} \times 100 \cr & = \frac{{\sqrt 3 }}{8}\pi \times 100 \cr & = 68\% \cr} $$
Hence, the free space in $$bcc$$  unit cell $$ = 100 - 68 = 32\% $$

$$\frac{{{}^rC{H_4}}}{{{r_x}}} = 2 = \sqrt {\frac{{{M_x}}}{{{M_{C{H_4}}}}}} = \sqrt {\frac{{{M_x}}}{{16}}} ,\,\,or\,\,{M_x} = 64$$

This figure clearly helps to establish the relation that force is directly proportional to area of contact and velocity gradient.
$$\eqalign{ & f \propto A,f \propto \frac{{du}}{{dz}} \cr & \therefore f \propto A\frac{{du}}{{dz}};f = \eta A\frac{{du}}{{dz}} \cr} $$
Where, $$\eta = $$  coefficient of viscosity.
$$A = $$  Area of contact.
$$\frac{{du}}{{dz}} = $$  veIocity gradient.

$$\eqalign{ & {\text{Mass of }}1{\text{ }}L{\text{ of vapour = volume}}\,\, \times {\text{ density}} \cr & {\text{ = 1000}} \times {\text{0}}{\text{.0006 = 0}}{\text{.6}}\,g \cr & V\,{\text{of}}\,{\text{liquid}}\,{\text{water}} = \frac{{mass}}{{density}} = \frac{{0.6}}{1} = 0.6c{m^3} \cr} $$

The beans are cooked earlier in pressure cooker because boiling point decreases with increasing pressure.

A gas is a collection of tiny particles separated from one another by large empty space and moving rapidly at random in all the directions. In the course of their motion, they collide with one another and also with the walls of the container. Due to frequent collisions, speeds and direction of motion of molecules keeps on changing. Thus, all the molecules in a sample of a gas do not have same speeds.

According to kinetic theory the gas molecules are in a state of constant rapid motion in all possible directions colloiding in a random manner with one another and with the walls of the container and between two successive collisions molecules travel in a straight line path but show haphazard motion due to collisions.

Given, number of moles of hydrogen $$\left( {{n_{{H_2}}}} \right)$$  and that of oxygen $$\left( {{n_{{O_2}}}} \right)$$  are equal.
∴ We have, the relation between ratio of number of moles escaped and ratio of molecular mass.
$$\frac{{{n_{{O_2}}}}}{{{n_{{H_2}}}}} = \sqrt {\frac{{{M_{{H_2}}}}}{{{M_{{O_2}}}}}} $$
where, $$M$$ = Molecular mass of the molecule.
$$\eqalign{ & \Rightarrow \frac{{{n_{{O_2}}}}}{{{n_{{H_2}}}}} = \sqrt {\frac{2}{{32}}} \cr & \Rightarrow \frac{{{n_{{O_2}}}}}{{{n_{{H_2}}}}} = \sqrt {\frac{1}{{16}}} \cr & \Rightarrow \frac{{{n_{{O_2}}}}}{{0.5}} = \frac{1}{4} \cr & \Rightarrow {n_{{O_2}}} = \frac{{0.5}}{4} = \frac{1}{8} \cr} $$

In body centered cubic, each atom/ion has a coordination number of 8.

The value of universal gas constant can be expressed in different units and its value would depend only on the units of the measurement.