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

Let $$V$$ be the original volume of bubble. The final volume will be $$8$$ $$V.$$  Let $$p$$ be the atmospheric pressure and $${p_1}$$ the pressure at the bottom
$$\eqalign{ & \therefore \,\,p \times 8\,V = {p_1} \times V; \cr & 8p = {p_1} \cr} $$
$${p_1} = $$  atmospheric pressure + pressure due to water lake
$$ = p + 7p$$
The $$p = 10$$  $$m$$ high, the $$7p$$  will be $$= 70$$  $$m$$ high
So the depth $$= 70$$  $$m$$

The value of $$a$$  is a measure of the magnitude of the attractive forces between the molecules of the gas. Greater the value of $$'a’,$$  larger is the attractive intermolecular force between the gas molecules.
The value of $$b$$  related to the effective size of the gas molecules. It is also termed as excluded volume.
The gases with higher value of $$a$$ and lower value of $$b$$ are more liquefiable, hence for $$C{l_2}$$ $$“a”$$ should be greater than for $${C_2}{H_6}$$  but for it $$b$$ should be less than for $${C_2}{H_6}.$$

The temperature at which a real gas behaves like an ideal gas is called Boyle’s temperature or Boyle’s point.

$$HCP$$  is a closed packed arrangement, in which the unit cell is hexagonal and coordination number is 12.

$$PV = nRT,P = 1\,atm,T = 273\,K,$$       $$V = 2.24\,L,$$   $$R = 0.0821\,\,L\,\,atm\,\,{K^{ - 1}}\,\,mo{l^{ - 1}}$$
$$n = \frac{{1 \times 2.24}}{{0.0821 \times 273}} = 0.099\,mol$$

Gases can be liquefied by lowering the temperature and increasing the pressure. An ideal gas have no intermolecular force of attraction, so it cannot be liquefied by applying high pressure and decreasing temperature.

Gases show deviation from ideal gas behaviour when the temperature is low and pressure is high. At low temperature, the volume of one molecule is not negligible in comparison to total volume and intermolecular force of attraction is maximum at low temperature and high pressure.

$$\eqalign{ & {n_{{N_2}}} = 53.57\,mole \cr & \frac{{{P_1}}}{{{P_2}}} = \frac{{{n_1}}}{{{n_2}}} \cr & {n_2} = \frac{{1.8}}{{200}} = 0.009 \times 53.57 = 0.48213 \cr & {n_{{N_2}}}\,{\text{escape}} = 53.087 \cr & {\text{Volume}} = \frac{{nRT}}{P} = 1264\,L \cr} $$

$$\eqalign{ & {\text{Mole fraction of}} \cr & {O_2} = \frac{{\frac{w}{{32}}}}{{\frac{w}{{32}} + \frac{w}{4}}} \cr & \,\,\,\,\,\,\, = \frac{w}{{32}} \times \frac{{32}}{{9w}} \cr & \,\,\,\,\,\,\, = \frac{1}{9} \cr & {\text{Partial pressure of oxygen}} \cr & = {P_{total}} \times {x_{{O_2}}} \cr & = {P_{total}} \times \frac{1}{9}\,\,{\text{or}}\,\,\frac{1}{9}{P_{total}} \cr} $$

TIPS/Formulae:
$$\eqalign{ & {U_{rms}} = \sqrt {\frac{{3RT}}{M}} \Rightarrow \sqrt {\frac{{3R{T_{{H_2}}}}}{2}} = \sqrt 7 \sqrt {\frac{{3R{T_{{N_2}}}}}{{28}}} ; \cr & \therefore \,\,{T_{{N_2}}} = 2{T_{{H_2}}}\,\,\,or\,\,\,{T_{{N_2}}} > {T_{{H_2}}} \cr} $$