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

$$Z = 1$$   under all conditions for an ideal gas.

Greater the polarisability of the interacting particles, greater is the magnitude of interaction energy.

$$\eqalign{ & {\text{Applying}}\,\,PV = \frac{m}{M}RT \cr & P{o_2} = \frac{4}{{32}} \times \frac{{0.0821 \times 273}}{1} \cr & \,\,\,\,\,\,\,\,\,\,\, = 2.8\,atm \cr & {P_{{H_2}}} = \frac{2}{2} \times \frac{{0.0821 \times 273}}{1} \cr & \,\,\,\,\,\,\,\,\,\, = 22.4\,atm \cr & {P_{{\text{total}}}} = {P_{{O_2}}} + {P_{{H_2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 2.8 + 22.4 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 25.2\,atm \cr} $$

In $$Ca{F_2}$$  (Fluorite structure), $$C{a^{2 + }}\,ions$$   are arranged in $$ccp$$  arrangement ( $$C{a^{2 + }}\,ions$$   are present at all corners and at the centre of each face of the cube) while $${F^ - }\,ions$$   occupy all the tetrahedral sites.

From the above figure, you can clearly see that coordination number of $${{F^ - }}$$ is 4 while that of $$C{a^{2 + }}$$  is 8.

$$\eqalign{ & {\text{According to ideal gas equation,}} \cr & pV = nRT \cr & \,\,\,\,p = \frac{n}{V}RT \cr & 1atm = 1mol\,{L^{ - 1}} \times 0.082 \times T \cr & T = \frac{1}{{0.082}} \cr & \,\,\,\,\, = 12K \cr} $$

$$\eqalign{ & C{O_2} + C \to 2CO \cr & {\text{Stoichoimetry ratio is 1 : 2}} \cr} $$
$$AT\,STP,\,P = 1\,atm,\,T = 273\,K,$$       $$R = 0.0821$$
Initial moles of $$C{O_2};n\left( {C{O_2}\,{\text{initial}}} \right) = \frac{{PV}}{{RT}}$$
$$\eqalign{ & = \frac{{1 \times 0.5}}{{0.0821 \times 273}} \cr & = 0.022\,mole \cr} $$
In final mixture no. of $$moles;\,n$$  ( $$C{O_2}/CO$$   mixture )
$$\eqalign{ & = \frac{{1 \times 0.7}}{{0.0821 \times 273}} \cr & = 0.031 \cr} $$
Increase in volume is by $$ = 0.031 - 0.022 = 0.009\,mole$$       of gas
Final no. of $$moles$$  of $$CO$$  i.e. $${n_{\left( {CO\,{\text{final}}} \right)}}$$
$$\eqalign{ & {n_{\left( {CO\,{\text{final}}} \right)}} = 2{n_{\left( {C{O_2}\,{\text{initial}}} \right)}} - {n_{\left( {C{O_2}\,{\text{final}}} \right)}} \cr & = 2\left( {0.022 - {n_{\left( {C{O_2}\,{\text{final}}} \right)}}\,\,\,\,\,\,\,\,...\left( {\text{i}} \right)} \right. \cr & {n_{\left( {CO\,\,{\text{final}}} \right)}} = 0.044 - 2{n_{\left( {C{O_2}\,{\text{final}}} \right)}}\,\,\,...\left( {{\text{ii}}} \right) \cr & \therefore {\text{NOW,}}\,{n_{\left( {CO\,\,{\text{final}}} \right)}} + {n_{\left( {CO\,\,{\text{final}}} \right)}} = 0.031 \cr & {n_{\left( {C{O_2}\,{\text{final}}} \right)}} = 0.031 - {n_{\left( {CO\,\,{\text{final}}} \right)}}\,\,\,\,\,...\left( {{\text{iii}}} \right) \cr & {\text{Substituting (ii) in eq}}{\text{. (i)}} \cr & {n_{\left( {CO\,\,{\text{final}}} \right)}} = 0.044 - 2\left[ {0.031 - {n_{\left( {CO\,\,{\text{final}}} \right)}}} \right] \cr & {n_{\left( {CO\,\,{\text{final}}} \right)}} = 0.044 - 0.062 + 2{n_{\left( {CO\,\,{\text{final}}} \right)}} \cr & {n_{\left( {CO\,\,{\text{final}}} \right)}} = 0.018\,mol. \cr & {\text{Volume of}} \cr & CO = V = \frac{{nRT}}{P} \cr & = \frac{{0.018 \times 0.0821 \times 273}}{1} \cr & = 0.40\,{\text{Litre}} \cr} $$
and volume of $$C{O_2} = 0.7\,litre - 0.4\,litre = 0.3\,litre$$
$$\therefore \,\,C{O_2} = 300\,mL,\,CO = 400\,mL$$

$$\eqalign{ & {\text{According to ideal gas equation,}} \cr & {p_1}V = {n_1}RT\,\,\,...({\text{i}}) \cr & {p_2}V = {n_2}RT\,\,...({\text{ii}}) \cr & {\text{From Eqs}}{\text{. (i) and (ii)}} \cr & \frac{{{p_1}V}}{{{p_2}V}} = \frac{{{n_1}RT}}{{{n_2}RT}} \Rightarrow \frac{{{p_1}}}{{{p_2}}} = \frac{{{n_1}}}{{{n_2}}} \cr & \frac{{{n_1}}}{{{n_2}}} = \frac{{{p_1}}}{{{p_2}}} = \frac{{170}}{{570}} = 0.30 \cr} $$

$$\eqalign{ & K = \frac{3}{2}{n_T}RT \Rightarrow {n_T} = \frac{{3000 \times 2}}{{3 \times 250 \times 8.314}} \cr & 3 \times {10^3} = \frac{3}{2}{n_T} \times 8.314 \times 250 \cr & \frac{x}{{20}} + \frac{{30 - x}}{{40}} = \frac{2}{{8.314}} \cr & \% \,Ne = 28.3 \cr} $$

Mass of gas can be determined by weighing a container in which it is enclosed as follows:
Mass of the gas = mass of the cylinder including gas $$ - $$ mass of empty cylinder
So, it is a wrong statement.

As we know that the van der Waals’ equation is
$$\left( {p + \frac{a}{{{V^2}}}} \right)\left( {V - b} \right) = RT$$
The real gases show ideal behaviour when pressure approaches zero and temperature is high. At this condition there is no force of attraction and repulsion between the molecules of gas.
Thus, the effect of $${\frac{a}{{{V^2}}}}$$  and $$b$$ is negligible, i.e.
$$\eqalign{ & pV = RT \cr & \frac{{pV}}{{RT}} = 1 \cr} $$
We also know $$\frac{{pV}}{{RT}} = Z$$   ( for ideal gas $$Z = 1$$   )
( $$Z$$ is compressibility factor )
Therefore an real gas behaves like ideal gas when the temperature is high and pressure is low.