States of Matter Solid, Liquid and Gas MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & {\text{TIPS/Formulae:}} \cr & {\text{Mole fraction of }}{O_2} = \frac{{{\text{Moles of}}\,{O_2}}}{{{\text{Total moles}}}} \cr & {\text{Partial pressure of}}\,{O_2} = \,{\text{Mole fraction of}}\,{O_2} \cr & {\text{Mole fraction of}}\,{O_2} = \frac{{\frac{W}{{32}}}}{{\frac{W}{{16}} + \frac{W}{{32}}}} = \frac{1}{3} \cr} $$

$$\eqalign{ & {\text{Most probable speed}}\,\left( {{C^ * }} \right) = \sqrt {\frac{{2RT}}{M}} \cr & {\text{Average Speed}}\,\left( {\bar C} \right) = \sqrt {\frac{{8RT}}{{\pi M}}} \cr & {\text{Root mean square velocity}}\,\left( c \right) = \sqrt {\frac{{3RT}}{M}} \cr & {C^ * }:\bar C:C = \sqrt {\frac{{2RT}}{M}} :\sqrt {\frac{{8RT}}{{\pi M}}} :\sqrt {\frac{{3RT}}{M}} \cr & = 1:\sqrt {\frac{4}{\pi }} :\sqrt {\frac{3}{2}} = 1:1.128:1.225 \cr} $$

According to Graham's law of diffusion,
$$\eqalign{ & {\text{rate of diffusion}} \cr & r \propto \frac{1}{{\sqrt M }},\,r \propto \frac{V}{t} \cr} $$
where, $$V$$ is the volume of the gas diffused in time $$t.$$
$$\eqalign{ & \frac{{{r_A}}}{{{r_B}}} = \sqrt {\frac{{{M_B}}}{{{M_A}}}} \,\,\,{\text{or}}\,\,\,\frac{{{V_A}}}{{{t_A}}} \times \frac{{{t_B}}}{{{V_B}}} = \sqrt {\frac{{{M_B}}}{{{M_A}}}} \cr & {\text{Given,}}\,\,{V_A} = {V_B} \cr & \therefore \,\,\,\,\,\,\,\,\,\,\,\frac{{10}}{{20}} = \sqrt {\frac{{{M_B}}}{{49}}} \Rightarrow \frac{1}{4} = \frac{{{M_B}}}{{49}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,{M_B} = \frac{{49}}{4} = 12.25\,u \cr} $$

Pressure exerted by hydrogen will be proportional to its mole fraction.
$${\text{Mole fraction of}}\,{H_2} = \frac{{\frac{W}{2}}}{{\frac{W}{{16}} + \frac{W}{2}}} = \frac{8}{9}$$

$${\text{van der Waals equation is}}$$     $$\left( {P + \frac{a}{{{V^2}}}} \right)\left( {V - b} \right) = RT$$
$$\eqalign{ & {\text{As given that }}b = 0; \cr & PV + \frac{a}{V} = RT \Rightarrow PV = RT - \frac{a}{V} \cr} $$
$${\text{Comparing with}}\,\,y = mx + c,$$       $${\text{Intercept}}\left( c \right) = RT,$$     $${\text{Slope}}\left( m \right) = - a$$
$$\eqalign{ & {\text{Slope}} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20.1 - 21.6}}{{3 - 2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 1.5. \cr & {\text{Thus, }}a{\text{ = }}1.5 \cr} $$

$$\eqalign{ & {\text{Average }}KE{\text{ }} = \,E\, = \frac{1}{2}M\,u_{rms}^2 \cr & \therefore \,\,u_{rms}^2 = \frac{{2E}}{M}\,{\text{or}}\,\,{u_{rms}} = \sqrt {\frac{{2E}}{M}} \cr} $$

In crystals the constituents ( $$atoms,$$  $$ions$$  or molecules) are arranged in definite orderly arrangement. When these crystals are cleaved they cut into regular patterns.

In general, the molar heat capacity for any process is given by
$$\eqalign{ & C = {C_v} + \frac{R}{{1 - n}},\,{\text{when}}\,P{V^n} = {\text{constant}} \cr & {\text{Here}}\,\,\frac{P}{V} = 1,\,\,\,i.e.\,\,P{V^{ - 1}} = {\text{constant}} \cr & {\text{For}}\,{\text{monoatomic}}\,{\text{gas,}}\,\,{C_v} = \frac{3}{2}R \cr & \therefore \,\,C = \frac{3}{2}R + \frac{R}{{1 - \left( { - 1} \right)}} = \frac{3}{2}R + \frac{R}{2} = \frac{{4R}}{2} = 2R. \cr} $$

More viscous liquids flow slowly.

Viscosity of liquids decreases as the temperature rises because at high temperature, molecules have high kinetic energy and can overcome the intermolecular forces to slip past one another between the layers.