NCERT Solutions for Class 11 Science Chemistry Chapter 5 - States Of Matter

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Answer:

At high altitudes, pressure is low. Hence, boiling takes place at lower temperature and therefore, cooking takes more time. In pressure cooker, pressure is high and hence boiling point increases.

Hence, the correct answer is option C.

Answer:

Due to surface tension, the surface of the water drops is under tension and tends to take spherical shape to reduce the tension.


Hence, the  correct answer is option B.

Answer:

According to Boyle's law, at constant temperature, the pressure of a fixed amount (number of moles, n) of a gas is inversely proportional to its volume.
p₁V₁ = p₂V₂ = p₃V₃ = p₄V₄ = Constant
Since, V₁ > V₂ > V₃ > V₄ 
Therefore, p₁ < p₂ < p₃ < p₄

Hence, the correct answer is option C.

Answer:

London dispersion forces operate only over a very short distance. The energy of interaction varies as inversely proportional to distance between two interacting particles to the power of 6. Large or more complex the molecules, greater is the magnitude of London forces. This is obviously due to the fact that the large electron clouds are easily distorted or polarised. Hence, greater the polarisability of interacting particles. Greater is the magnitude of interaction energy.

Hence, the correct answer is option C.

Answer:

Partial charge is a small charge developed by displacement of electrons. It is less than unit electronic charge and is represented as δ⁺ or δ^–.

Hence, the correct answer is option C.

Answer:

According to dalton's law, the partial pressure of oxygen is given as follows:

${\mathrm{p}}_{{\mathrm{O}}_2}={\mathrm{x}}_{{\mathrm{O}}_2}\times {\mathrm{P}}_{\mathrm{total}}$

Since, there is a mixture of 1:4 dihydrogen and dioxygen, therefore, the  mole fraction of oxygen be given as follows:

${\mathrm{x}}_{{\mathrm{O}}_2}=\frac{4}{5}$

Also, it's given that the the total pressure is 1 atm.  

Therefore, ${\mathrm{p}}_{{\mathrm{O}}_2}= \frac{4 }{5 }\times 1=0.8 \mathrm{atm}$

$(\because 1 \mathrm{atm}=1.01325\times {10}^5\mathrm{Pa} \mathrm{or} \mathrm{N} {\mathrm{m}}^{-2} )$

Therefore, ${\mathrm{p}}_{{\mathrm{O}}_2}=0.8\times {10}^5=8\times {10}^4{\mathrm{Nm}}^{-2}$

Hence, the correct  answer is option C.

Answer:

Gay Lussac's law states that at constant volume, pressure of a fixed amount of a gas varies directly with the temperature.
 

Answer:

Critical temperature of a gas may be defined as that temperature above which it cannot be liquified howsoever high pressure may be applied on the gas. Higher the critical temperature, more easily is the gas liquified. Hence, order of liquefaction starting with the gas liuefying first will be O₂, N₂, H₂, He.

Hence, the correct answer is option D.O2,N2,H2,He

Answer:

Resistance to the flow of liquid is called viscosity.

F=η.A

Here,  is a velocity gradient, η is a viscosity coefficient, A is area

Therefore, η =.

Putting, F = N, dx=m, A= m²,v=ms^-1

η= N/m². m/ms^-1
η= Nsm^-2

Hence, the correct answer is option B.

Answer:

A liquid starts boiling when the pressure on its surface is equal to the atmopheric presssure. Thus liquid will boil first at Shimla because its atmospheric pressure is minimum.


Hence, the correct  answer is option A.

Answer:

If we plot a graph pV vs p of gases, at constant temperature, pV will be constant (Boyle’s law) and pV vs p graph at all pressures will be a straight line parallel to x-axis. This is the condition for ideal gas behaviour.

Hence, the correct answer is option A.

Answer:

Viscosity is a measure of resistance to flow which arises due to the internal friction between layers of fluid as they slip past one another while liquid flows. 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.

Hence, the correct answer is option C.

Answer:

Increase in temperature increases the kinetic energy of the molecules and effectiveness of intermolecular attraction decreases, so surface tension decreases as the temperature is raised.

Hence, the correct answer is  option B.
 

Answer:

In gaseous state, molecules are in a state of random motion, i.e., it is the state in which molecules are disorderly arranged. Gaseous state has higher entropy than the liquid as well as solid.

Hence, the  correct answers are options B and C.

Answer:

1 mole of dioxygen represents 32 g of O₂, 22.7 L of O₂ or 6.022 x 10²³ molecules of O₂ gas.

Hence, the correct answers are options A and D.

Answer:

A gas which obeys the ideal gas equation, PV = nRT, under all conditions of temperature and pressure is called an 'ideal gas'.
The gases are found to obey the gas laws fairly well when the pressure is low or the temperature is high.
Such gases are, therefore, known as 'real gases'. All gases are real gases. Hence, at high pressure and low temperature, a real gas deviates most ideal behaviour.

Hence, the correct answers are options B and C.

Answer:

The vapour pressure of water doesn't depend upon the quantity of water and its volume, rather it changes with the addition of salt and decreasing the temperature.

Hence, the correct answers are options B and D. 
 

Answer:

Avogadro's law law states that volume is directly proportional to the number of moles of the gas at constant pressure and temperature.

$\mathrm{V}\propto \mathrm{n}$ ( at constant temperature and pressure)

Therefore, gases having maximum number of moles will be having maximum volume.

We know that $\mathrm{moles}=\frac{\mathrm{given} \mathrm{mass}}{\mathrm{molar} \mathrm{mass}}$

thus, $\mathrm{moles} \mathrm{of} \mathrm{CO} = \frac{1}{28}=0.0357$
$\mathrm{moles} \mathrm{of} {\mathrm{H}}_2\mathrm{O} =\frac{1}{18}=0.055$
$\mathrm{mles} \mathrm{of} {\mathrm{CH}}_4=\frac{1}{16}=0.0625$
$\mathrm{moles} \mathrm{of} \mathrm{NO}=\frac{1}{30}=0.033$

Therefore, the gas occupying maximum volume is CH₄ and minimum volume is occupied by NO.


 

Answer:

The physical properties of water are different in different states but the chemical composition of water in all the three states i.e. solid, liquid and gas remains the same.

Answer:

The different states of matter can be determined on the basis of temperature, pressure, mass and volume.

Answer:

(a) These hydrogen halides have 'dipole interactions' and 'london forces' which  act as intermolecular forces.

(b) The values inndicate that the london forces are pre-dominant. As the surface area increases on moving from fluorine to iodine, london forces increases and boiling point also increases.

(c) Fluorine is highly electronegative and therefore, HF includes hydrogen bonding among its molecules. As a result, have higher boiling point.

Answer:

STP is the condition where pressure is 1 atm and temperature is  273.15 K. Under these conditions the volume occupied by 1 mole of a gas is 22.4 L. Therefore, the volume occupied by  both  N₂ and Ar will be 22.4 L.

Answer:

Ideal gas is defined as the gas which don't  have attractive forces and follows all the laws under all the conditions of temperature and pressure. A real gas behaves ideally when the temperature is high and the pressure is low.

Answer:

Liquefaction of  gases is defined on the basis of their respective critical temperatures i.e.the temperature above which a gas can't be liquified however, larger the pressure may  be.
In the present case, the critical temperature of gas B is high and that of gas A is low. Thus, gas B  needs to  be  cooled even on application of pressure.

Answer:

Unit of R depends upon those units in which p, V and T are measured, $\mathrm{R}=\frac{\mathrm{PV}}{\mathrm{nT}}$

depending upon these variables, we have three different values of R as follows:

$$\begin{gathered} \mathrm{R}=8.314 \mathrm{J} {\mathrm{K}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{R}=0.0821 \mathrm{L} \mathrm{atm} {\mathrm{K}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{R}=2 \mathrm{Cal} {\mathrm{K}}^{-1}{\mathrm{mol}}^{-1} \end{gathered}$$

Answer:

The statement is true under low pressure and high temperature only. If this condition is followed at all the temperatures and pressures, then an ideal gas can't be  liquified due to absence of force of attractions between the molecules.

Answer:

Hexane being non-polar molecule will have the least attractive forces i.e. london forces.
Both water and alcohol involves dipole interactions as well as hydrogen bonding bu the extent of hydrogen bonding in water is more than alcohol. Therefore, the surface tension of water is greater than alcohol.

Thus,  the  correct  order is:

hexane < alcohol < water

Answer:

In order to obtain the total pressure of a dry gas (free from water vapours), the aqueous tension must be  subtracted.


${\mathrm{P}}_{\mathrm{dry} \mathrm{gas}}={\mathrm{P}}_{\mathrm{moist} \mathrm{gas}}-\mathrm{aqueous} \mathrm{tension}$.

Answer:

The energy which arises due to motion of atoms or molecules in a body is thermal energy which is measured as average kinetic energy of molecules.
The average kinetic energy is directly proportional to the temperature. As the temperature increases, the average kinetic energy of molevules increases.
 

Answer:

Since, fluorine is highly electronegative in nature, therefore, HF molecule  in liquid state possess dipolar interactions.

Answer:

The assumption is applicable to gases under low pressure and high temperature only i.e. under ideal conditions. But real gases do contain attractive forces among their molecules and therefore, can be easily liquefied as compared to ideal gas.

Answer:

(i) The value of compressibility factor of an  ideal gas is 1.

(ii) Boyle's temperature is the temperature at  which real gas starts behaving like an ideal gas. Under such conditions, Z > 1.

Answer:

Critical temperature is the temperature above which a gas can't be liquefied. In this case, CO₂ can't be liquefied since the temperature is above the critical temperature.

Answer:

(i) Molar volume is the excluded volume or co-volume and is directly  proportional to the size of the molecule. Thus, the value of 'b' increases in the order: He < H₂ < O₂ < CO₂.
(ii) Van der Waal's constant 'a' is the measure of magnitude of inter-molecular attraction. The magnitude of inter-molecular attractions increases with increase in size of electron cloud in a molecule. Hence for the given gases magnitude of 'a' decreases in the following order:
CH₄ >  O₂ > H₂.
 

Answer:

$p_{\mathrm{ideal}}=p_{\mathrm{real}}+\frac{a{n}^2}{V^2}$
On rearranging the equation, we get
$a=\frac{\left(p_{\mathrm{ideal}}-p_{\mathrm{real}}\right)V^2}{n^2}$
So, the unit of ‘a’ can be calculated as follows:
$$\begin{gathered} a=\frac{{\mathrm{Nm}}^{-2}{\left({\mathrm{m}}^3\right)}^2}{{\left(\mathrm{mol}\right)}^2} \mathrm{and} a=\frac{\mathrm{atm}{\left({\mathrm{dm}}^3\right)}^2}{{\left(\mathrm{mol}\right)}^2} \\ a={\mathrm{Nm}}^4{\mathrm{mol}}^{-2} \mathrm{and} a=\mathrm{atm} {\mathrm{dm}}^6{\mathrm{mol}}^{-2} \end{gathered}$$

Answer:

The phenomenons that can be explained on the basis of surface tension are as follows:

(i) Spherical shape of a liquid drop.

(ii) Rise and fall of liquid in a capillary i.e. capillary action.

Answer:

Hexane is non-polar and thus, have london forces only. In case of water and glycerol, hydrogen bonding is present due to polar (O-H) bonds. The extent of hydrogen bonding is higher in case of glycerol due to the presence of three (O-H) bonds.

Thus, the correct order of viscosity is as follows: hexane < water < glycerol.
 

Answer:

Increase in temperature decreases the intermolecular forces because the kinetic energy of the molecules increases. As the temperature increases, the kinetic energy of molecules surpasses intermolecular forces and therefore, the viscosity of a liquid decreases.

Answer:

(i) The volume of a gas will decrease if its pressure is increased at constant temperature because according to Boyle’s law, at constant temperature, the pressure of a fixed amount (i.e., number of moles n) of gas varies inversely with its volume.


(ii) The volume of a gas will increase by two times if the temperature is increased from 200K to 400K at a constant pressure because according to Charles’ law, pressure remaining constant, the volume of a fixed mass of a gas is directly proportional to its absolute temperature.

Answer:

(i) At low pressures, measured and calculated volumes approach each other i.e. real gas approaches ideal gas behaviour.

(ii) At high pressures, the measured volume is more than the calculated volume i.e. real gas deviates largely from ideal gas behaviour.

(iii) At the intersection of two curves, real gas behaves as an ideal gas.

 

Answer:

(i) The graph at constant molar volume is called isochore.

(ii) The graph at constant temperature is called isotherm.

(iii) The graph at constant pressure is called isobar.

Therefore, the correct matching may be done as follows:

(i)$\to$(c) ; (ii)$\to$(a) ; (iii)$\to$(d)

Answer:

(i) Boyle's law states that at constant temperature, the pressure of gas is inversely proportional to the temperature.

(ii) Charle's law states that at constant pressure, the volume of gas is directly proportional to the temperature.

(iii) Dalton's law states that at constant temperature and volume, the total pressure of all the gases in a mixture is equal to the sum of their respective partial pressures.

(iv) Avogadro's law states that at constant temperature an pressure, equal volume of all that gases contain equal number of molecules.

Therefore, the correct matching may be done as follows:

(i)$\to$(e) ; (ii)$\to$(d) ; (iii)$\to$(b) ; (iv)$\to$(a)
 

Answer:

	Graphical representation		x and y co-ordinates
(i)		(b)	p vs. V
(ii)		(c)	$p vs. \frac{1}{V}$
(iii)		(a)	pV vs. V

Answer:

Intermolecular forces and thermal energy both decides the state of matter and thus, a balance should be maintained among the two.

Hence, the correct answer is  option A.

Answer:

At constant temperature, pV vs V plot for real gases is not a straight line because their molecules do contain intermolecular forces.

Hence, the  the correct  answer is option B.
 

Answer:

boiling point is the temperature at which vapour pressure of a liquid is equal to the external pressure. The vapour pressure decreases with altitude.

Hence, the  correct answer is option C.

Answer:

The gases can't be liquefied above the critical temperature, however, larger the pressure may be. This is because of the higher molecular speed and poor intermolecular attractions.

Hence, the correct answer is option A.

Answer:

At critical temperature, the density of liquid becomes equal to liquid phase and it passes continuously to the gaseous state.

Hence, the correct  answer is option A.

Answer:

Liquids always try to have lesser surface area so that the surface tension may be reduced and due to this reason, the rop of liquid tend to occupy spherical shape.

Hence, the correct  answer is option D.

 

Answer:

(i) Carbon dioxide will exist in gaseous state between the points a and b at temperature T₁.

(ii) At point 'b' CO₂ will start liquefying when temperature is T₁.

(iii) At point 'g' CO₂ will be completely liquefied when temperature is T₂.

(iv) No, condensation will not take place when the temperature is T₃ because the temperature T₃ is greater than critical temperature (the highest temperature at which liquid carbon dioxide is observed).

(v) The portion between points 'b' and 'c' of the isotherm at T₁ represent liquid and gaseous CO₂ at equilibrium.



 

Answer:

(i) Boiling point of A = approximately 315 K, B = approximately 345 K

(ii) If we take liquid C in a closed vessel and heat it continuously, it will not boil because vapour pressure will keep on increasing.

(iii) The boiling point of liquid is the temperature at which vapour pressure becomes equal to atmospheric pressure. At high altitude, atmospheric pressure is low (say 60 mm Hg). So, liquid D will boil at temperature of approximately 313 K corresponding to this pressure.

(iv) At high altitudes atmospheric pressure is low. The boiling point of liquid is the temperature at which vapour pressure becomes equal to atmospheric pressure. Therefore, liquids at high altitudes boil at lower temperatures in comparison to that at sea level. Since water boils at low temperature on hills, the pressure cooker is used for cooking food.

Answer:

At the critical temperature, the density of liquid becomes equal to the vapour and therefore, the boundary of separation disappeared and in this situation the substance is called supercritical fluid.

Answer:

On heating, glass melts, and the surface of the liquid tends to take the rounded shape at the edges which have minimum surface area. This is called fire polishing of glass.

Answer:

When a liquid flows over a fixed surface, the layer of molecules in the immediate contact of surface is stationary. The velocity of the upper layers increases as the distance of layers from the fixed layer increases. This type of flow in which there is a regular gradation of velocity on passing from one layer to the next is called laminar flow.

In laminar flow, the velocity of molecules is not same  in all the layers because of the friction exerted by the layer to the layer next to it.
 

Answer:

The path is as follows:
We can move from point A to F vertically by increasing the temperature, then we can reach the point G by compressing the gas at the constant temperature along this isotherm (isotherm at 31.1°C). The pressure will increase. Now we can move vertically down towards D by lowering the temperature. As soon as we cross the point H on the critical isotherm we get liquid. We end up with liquid but in this series of changes we do not pass through two-phase region. If process is carried out at the critical temperature, substance always remains in one phase.