States of Matter Solid, Liquid and Gas MCQ Questions & Answers in Physical Chemistry | Chemistry
Learn States of Matter Solid, Liquid and Gas MCQ questions & answers in Physical Chemistry are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
151.
van der Waal’s equation for a gas is stated as, $$P = \frac{{nRT}}{{V - nb}} - a{\left( {\frac{n}{V}} \right)^2}.$$
This equation reduces to the perfect gas equation, $$P = \frac{{nRT}}{V}$$ when,
A
temperature is sufficient high and pressure is low.
B
temperature is sufficient low and pressure is high.
C
both temperature and pressure are very high.
D
both temperature and pressure are very low.
Answer :
temperature is sufficient high and pressure is low.
$$\eqalign{
& {\text{Given}}\,\,P = \frac{{nRT}}{{v - nb}} - a{\left( {\frac{n}{v}} \right)^2} \cr
& {\text{Which can also be written as}} \cr
& \left[ {P + \frac{{{n^2}a}}{{{V^2}}}} \right]\left( {V - nb} \right) = nRT \cr} $$
At low pressure and high temperature the effect of $$\frac{a}{{{V^2}}}$$ and $$b$$ is negligible hence $$PV = nRT.$$
152.
The relations between various variables of gaseous substances are given along with their formulae. Mark the incorrect relationship.
A
Density and molar mass $$:M = \frac{{dRT}}{P}$$
B
Universal gas constant, $$P,V,T:R = \frac{{PV}}{{nT}}$$
C
Volume and pressure $$:{V_2} = \frac{{{P_2}{V_1}}}{{{P_1}}}$$
D
Volume and temperature $$:{V_2} = \frac{{{V_1}{T_2}}}{{{T_1}}}$$
Answer :
Volume and pressure $$:{V_2} = \frac{{{P_2}{V_1}}}{{{P_1}}}$$
153.
At $${25^ \circ }C$$ and $$730\,mm$$ pressure, $$380\,mL$$ of dry oxygen was collected. If the temperature is constant, what volume will the oxygen occupy at $$760\,mm$$ pressure?
A
365$$\,mL$$
B
2$$\,mL$$
C
10$$\,mL$$
D
20$$\,mL$$
Answer :
365$$\,mL$$
As the temperature is constant, Boyle’s law is applicable.
$$\eqalign{
& {p_1}{V_1} = {p_2}{V_2} \cr
& {V_1} = 380\,mL,\,{p_1} = 730\,mm,\,{V_2} = ? \cr
& {p_2} = 760\,mm \cr
& 730 \times 380 = 760 \times {V_2} \cr
& {V_2} = \frac{{730 \times 380}}{{760}} = 365\,mL \cr} $$
154.
$$X$$ and $$Y$$ are two volatile liquids with molar weights of $$10\,g\,mo{l^{ - 1}}$$ and $$40\,g\,mo{l^{ - 1}}$$ respectively. Two cotton plugs, one soaked in $$X$$ and the other soaked in $$Y,$$ are Simultaneously placed at the ends of a tube of length $$L = 24\,cm,$$ as shown in the figure. The tube is filled with an inert gas at 1atmosphere pressure and a temperature of $$300\,K.$$ Vapours of $$X$$ and $$Y$$ react to form a product which is first observed at a distance $$d\,cm$$ from the plug soaked in $$X.$$ Take $$X$$ and $$Y$$ to have equal molecular diameters and assume ideal behaviour for the inert gas and the vapours.
The value of $$d$$ in $$cm$$ ( shown in the figure ), as estimated from Graham's law, is
A
8
B
12
C
16
D
20
Answer :
16
According to Graham's law, $$r \propto \frac{1}{{\sqrt M }}$$
As all conditions are identical for $$X$$ and $$Y,$$
$$\eqalign{
& \frac{{{r_X}}}{{{r_Y}}} = \sqrt {\frac{{{M_Y}}}{{{M_X}}}} \Rightarrow \frac{d}{{24 - d}} = \sqrt {\frac{{40}}{{10}}} = 2 \cr
& d = 48 - 2d \Rightarrow 3d = 48 \Rightarrow d = 16\,cm \cr} $$
155.
A manifestation of surface tension is :
A
rise of liquid in a capillary tube
B
spherical shape of liquid drops
C
upward movement of water in soils
D
All the above
Answer :
All the above
All the given phenomenon occurs due to surface tension
156.
There are three closed containers in which equal amount of the gas are filled.
If all the containers are placed at the same temperatures, then find the incorrect options –
A
Pressure of the gas is minimum in (III) container
B
Pressure of the gas is equal in (I) and (II) container
C
Pressure of the gas is maximum in (I)
D
The ratio of pressure in (II) and (III) container is 4 : 3
Answer :
Pressure of the gas is equal in (I) and (II) container
$$\eqalign{
& n,T\,{\text{same hence}}\,P \propto \frac{1}{V}, \cr
& {V_1} = 1000\,c{m^3} \cr
& {V_2} = \pi {\left( {10} \right)^2} \times 10 = 1000\,\pi \,c{m^3} \cr
& {V_3} = \frac{4}{3}\pi {\left( {10} \right)^3} = \frac{4}{3}\pi \,1000\,c{m^3} \cr} $$
∴ Pressure of the gas is minimum in (III) container, Pressure of the gas is maximum in (I),
157.
A given metal crystallises out with a cubic structure having edge length of $$361\,pm.$$ If there are four metal atoms in one unit cell, what is the radius of one atom?
A
40$$\,pm$$
B
127$$\,pm$$
C
80$$\,pm$$
D
108$$\,pm$$
Answer :
127$$\,pm$$
Given, edge length $$= 361\,pm$$
Four metal atoms in one unit cell
i.e. effective number in unit cell $$(z)$$ $$= 4$$ (given)
It is a $$FCC$$ structure
$$\eqalign{
& \therefore \,\,\,{\text{Face diagonal}} = 4r \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt 2 a = 4r \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,r = \frac{{\sqrt 2 \times 361}}{4} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 127\,pm \cr} $$
158.
The ionic radii of $${A^ + }$$ and $${B^ - }\,ions$$ are $$0.98 \times {10^{ - 10}}m$$ and $$1.81 \times {10^{ - 10}}m.$$ The coordination number of each ion in $$AB$$ is
A
4
B
8
C
2
D
6
Answer :
6
Given, ionic radius of cation $$\left( {{A^ + }} \right) = 0.98 \times {10^{ - 10}}m$$
Ionic radius of anion $$\left( {{B^ - }} \right) = 1.81 \times {10^{ - 10}}m$$
∴ Coordination number of each ion in $$AB$$ = ?
Now, we have
$$\eqalign{
& {\text{Radius ratio}} = \frac{{{\text{Radius of cation}}}}{{{\text{Radius of anion}}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.98 \times {{10}^{ - 10}}m}}{{1.81 \times {{10}^{ - 10}}m}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.541 \cr} $$
If radius ratio range is in between $$0.441 - 0.732,\,ion$$ would have octahedral structure with coordination
number $$‘six’.$$
159.
The term that corrects for the attractive forces present in a real gas in the van der Waals equation is
A
$$nb$$
B
$$\frac{{a{n^2}}}{{{V^2}}}$$
C
$$ - \frac{{a{n^2}}}{{{V^2}}}$$
D
$$ - nb$$
Answer :
$$\frac{{a{n^2}}}{{{V^2}}}$$
Correction factor for attractive force for n moles of real gas is given by the term mentioned in (B).
160.
The pressure of a 1 : 4 mixture of dihydrogen and dioxygen enclosed in a vessel is one atmosphere. What would be the partial pressure of dioxygen ?