# NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter

NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter: Temperature is a relative measure of indication of coldness (or hotness) while heat is the form of energy transferred between two or more systems or a system and its surroundings due to the temperature difference. The solutions of NCERT class 11 physics chapter 11 thermal properties of matter start with the questions related to temperature conversions from one scale to another. All the questions discussed in the CBSE NCERT solutions for class 11 physics chapter 11 thermal properties of matter are important to understand the concept studied in the chapter. NCERT solutions are an important tool to perform well in exams. Consider a day in the winter season. Let the temperature at 6 am is 5 degree and 9 am be 10 degrees. We will say that 6 am is much colder than 9 am or we can say 9 am is hotter than 6 am which means cold and hot are relative terms.  We know that a glass of ice left on a table on a hot day eventually warms up whereas a cup of hot coffee on the same table cools down. It means that when the temperature of the body and its surrounding medium are different, heat transfer takes place between the system and the surrounding medium until the body and the surrounding medium are at the same temperature. So hot coffee becomes cold and cold water becomes warmer. The questions are asked in two sessions which are:

Exercise

Questions based on the following topics are discussed in the NCERT solutions for class 11 physics chapter 11 thermal properties of matter.

11.2 Temperature and heat

11.3 Measurement of temperature

11.4 Ideal-gas equation and absolute temperature

11.5 Thermal expansion

11.6 Specific heat capacity

11.7 Calorimetry

11.8 Change of state

11.9 Heat transfer

11.10 Newton’s law of cooling

The difference between temperature and heat can be well understood from the following graph.

Even though heat is increased during the phase change, the temperature remains constant until the phase change has occurred. That is increasing heat always does not imply that there is an increase in temperature.

## NCERT solutions for class 11 physics chapter 11 thermal properties of matter exercise

The relation between Kelvin and Celcius scale is TK = TC + 273.15

Triple Point of Neon in Kelvin TK = 24.57 K

Triple Point of Neon in Celcius TC = TK -273.15 = 24.57 -273.15 =-248.58 oC

Triple Point of carbon dioxide in Kelvin TK = 216.55 K

Triple Point of carbon dioxide in Celcius T= TK -273.15 = 216.55 - 273.15 = -56.60 oC

The relation between Celcius and Fahrenheit scale is $T_{F}=\frac{9}{5}T_{C} + 32$

Triple Point of Neon in Fahrenheit is

$T_{F}=\frac{9}{5}\times (-248.58) + 32$

TF = -415.44 oC

Triple Point of carbon dioxide in Fahrenheit is

$T_{F}=\frac{9}{5}\times (-56.60) + 32$

TF = -69.88 oC

200 A = 273 K

$T_{K}=\frac{273}{200}T_{A}$

350 B = 273 K

$T_{K}=\frac{273}{350}T_{B}$

Equating TK From the above two equations we have

$\\\frac{273}{200}T_{A}=\frac{273}{350}T_{B}\\ T_{A}=\frac{4}{7}T_{B}$

$\inline R=R_{0}\left [ 1+\alpha (T-T_{0}) \right ]$

$\inline R=R_{0}\left [ 1+\alpha (T-T_{0}) \right ]$

R0 = $\inline 101.6\Omega$    T0 = $273.16 K$

R = $\inline 165.5\; \Omega$    R = $600.5 K$

Putting the above values in the given equation we have

$\\\alpha =\frac{165.5-101.6}{600.5-27316}\\ \alpha =1.92\times 10^{-2}\ K^{-1}$

For R = $\inline 123.4\; \Omega$

$\\T=T_{0}+\frac{1}{\alpha }\left ( \frac{R}{R_{0}}-1 \right )\\ T=273.16+\frac{1}{1.92\times 10^{-3}}\left ( \frac{123.4}{101.6}-1 \right )\\ T=384.75K$

Unlike the melting point of ice and boiling point of water, the triple point of water has a fixed value of 273.16 K. The melting point of ice and boiling point of water vary with pressure.

Q. 11.4 (d) Answer the following :

(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?

Let at a certain temperature the reading on Fahrenheit and Kelvin Scale be T and TK respectively

$T_{F}-32=\frac{9}{5}(T_{K}-273)$                                $(i)$

Let at another temperature the reading on Fahrenheit and Kelvin Scale be T' and T'K respectively

$T'_{F}-32=\frac{9}{5}(T'_{K}-273)$                                 $(ii)$

Subtracting equation (ii) from (i)

$T_{F}-T'_{F}=\frac{9}{5}(T_{K}-T'_{K})$

For T- T'K = 1 K, TF - T'F = 9/5

Therefore corresponding to 273.16 K the absolute scale whose unit interval size is equal to that of the Fahrenheit scale

$\\T_{F}=\frac{9}{5}\times 273.16\\ T_{F}=491.688$

## Q. 11.5 (a) Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :

 Temperature Pressure Thermometer A Pressure Thermometer B Triple-point of water $\inline 1.250\times 10^{5}Pa$ $\inline 0.200\times 10^{5}Pa$ Normal melting point of sulphur $\inline 1.797\times 10^{5}Pa$ $\inline 0.287\times 10^{5}Pa$

(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?

$\\PV=nRT\\$

$\frac{P}{T}=\frac{nR}{V}$

As the moles of oxygen and hydrogen inside the thermometers and the volume occupied by the gases remain constant P/T would remain constant.

The triple point of water(T1) = 273.16 K

Pressure in thermometer A at a temperature equal to the triple point of water (P1) = $\inline 1.250\times 10^{5}Pa$

Pressure in thermometer A at a temperature equal to Normal melting point of sulphur (P2) = $\inline 1.797\times 10^{5}Pa$

The normal melting point of sulphur as read by thermometer A T2 would be given as

$T_{2}=\frac{P_{2}T_{1}}{P_{1}}$

$\\T_{2}=\frac{1.797\times 10^{5}\times 273.16}{1.250\times 10^{5}}\\ T_{2}=392.69\ K$

Pressure in thermometer B at a temperature equal to the triple point of water (P'1) = $\inline 0.200\times 10^{5}Pa$

Pressure in thermometer B at a temperature equal to Normal melting point of sulphur (P'2) = $\inline 0.287\times 10^{5}Pa$

The normal melting point of sulphur as read by thermometer B T'2 would be given as

$\\T'_{2}=\frac{P'_{2}T_{1}}{P'_{1}}\\ T'_{2}=\frac{0.287\times 10^{5}\times 273.16}{0.200\times 10^{5}}\\ T'_{2}=391.98\ K$

 Temperature Pressure Thermometer A Pressure Thermometer B Triple-point of water $\inline 1.250\times 10^{5}Pa$ $\inline 0.200\times 10^{5}Pa$ Normal melting point of sulphur $\inline 1.797\times 10^{5}Pa$ $\inline 0.287\times 10^{5}Pa$

(b) What do you think is the reason behind the slight difference in answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?

The slight difference in answers of thermometers A and B occur because the gases used in the thermometers are not ideal gases. To reduce this discrepancy the experiments should be carried out at low pressures where the behaviour of real gases tend close to that of ideal gases.

At 27 oC the 63 cm (l1) mark on the steel tape would be measuring exactly 63 cm as the tape is calibrated at 27 oC

Coefficient of linear expansion of steel $\inline =1.20\times 10^{-5}K^{-1}.$

Actual length when the scale is giving a reading of 63 cm on at 45 oC is l2

$\\l_{2}=l_{1}(1+\alpha \Delta T)\\ =63\times (1+1.20\times 10^{-5}\times (45-27))\\ =63.013608cm$

The actual length of the steel rod on a day when the temperature is 45 oC is 63.013608 cm.

Length of the same steel rod on a day when the temperature is 63 cm.

Diameter of the steel shaft at 27 oC (T1) d1 = 8.70 cm

The diameter of the central hole in the wheel d2 = 8.69 cm

Coefficient of linear expansion of the steel $\inline \alpha _{steel}=1.20\times 10^{-5}K^{-1}$.

The wheel will slip on the shaft when the diameter of the steel shaft becomes equal to the diameter of the central hole in the wheel.

Let this happen at temperature T

$\\d_{2}=d_{1}(1+\alpha (T-T_{1})\)\\ 8.69=8.7(1+1.2\times 10^{5}(T-27))\\ T=-68.79\ ^{o}C$

Coefficient of linear expansion of copper $\alpha$$\inline =1.70\times 10^{-5}K^{-1}.$

Coefficient of superficial expansion of copper is $\beta$

$\\\beta =2\alpha \\ \beta =2\times 1.7\times 10^{-5}\\ \beta =3.4\times 10^{-5}K^{-1}$

Diameter of the hole at 27 oC (d1) = 4.24 cm

Area of the hole at 227 oC is

$\\A_{2}=A_{1}(1+\beta \Delta T)\\ A_{2}=\pi \left ( \frac{4.24}{2} \right )^{2}(1+3.4\times 10^{-5}(227-27))\\ A_{2}=14.215\ cm^{2}$

Let the diameter at 227 oC be d2

$\\\pi \left ( \frac{d_{2}}{4} \right )^{2}=14.215\\ d_{2}=4.254cm$

Change in diameter is d-d1 = 4.24 -4.254 = 0.014 cm.

Youngs Modulus of Brass, $Y=0.91\times 10^{11}$

Co-efficient of linear expansion of Brass, $\alpha =2.0\times 10^{-5}K^{-1}$

The diameter of the given brass wire, d = 2.0 mm

Length of the given brass wire, l = 1.8 m

Initial Temperature T1 = 27 oC

Final Temperature T2 = -39 o C

$\\Y=\frac{Stress}{Strain}\\ Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}\\ F=\frac{\Delta lAY}{l}\\ F=\frac{(l\alpha \Delta T)AY}{l}\\ F=\alpha \Delta TAY\\ F=\alpha \Delta TY\pi \frac{d^{2}}{4}\\ F=\frac{2.0\times 10^{-5}\times (-39-27)\times 0.91\times 10^{11}\times \pi \times (2\times 10^{-3})^{2}}{4}$

$F=-378N$

The tension developed in the wire is 378 N. The negative sign signifies this tension is inward.

Length of the rods l = 50 cm

Co-efficient of linear expansion of brass, $\\\alpha _{b}=2\times 10^{-5}K^{-1}$

Co-efficient of linear expansion of steel, $\\\alpha _{s}=1.2\times 10^{-5}K^{-1}$

Initial Temperature T1 = 40.0 oC

Final Temperature T2 = 250 oC

Change in length of brass rod is

$\\\Delta l_{t}=l_{t}\alpha_{b} \Delta T\\ \Delta l_{t}=50\times 2.0\times 10^{-5}\times (250-40)\\ \Delta l_{t}=0.21cm$

Change in length of the steel rod is

$\\\Delta l_{s}=l_{s}\alpha_{s} \Delta T\\ \Delta l_{s}=50\times 1.2\times 10^{-5}\times (250-40)\\ \Delta l_{s}=0.126cm$

Change in length of the combined rod is

$\\\Delta l=\Delta l_{s}+\Delta l_{b}\\ \Delta l=0.126+0.21\\ \Delta l=0.336cm$

## Q. 11.11 The coefficient of volume expansion of glycerine is $\inline 49\times 10^{-5}K^{-1}.$What is the fractional change in its density for a $\inline 30^{\circ}C$ rise in temperature?

Coefficient of volume expansion of glycerine is $\gamma =49\times 10^{-5}K^{-1}$

Let initial volume and mass of a certain amount of glycerine be V and m respectively.

Initial density is

$\rho =\frac{m}{V}$

Change in volume for a 30 oC rise in temperature will be

$\\\Delta V=V(\gamma \Delta T)\\ \Delta V=V(49\times 10^{-5}\times 30)\\ \Delta V=0.0147V$

Final Density is

$\\\rho' =\frac{m}{V+\Delta V}\\ \rho'=\frac{m}{1.0147V}\\ \rho'=\frac{0.986m}{V}$

Fractional Change in density is

$\\\frac{\rho'-\rho}{\rho}\\ =\frac{\frac{0.986m}{v}-\frac{m}{v}}{\frac{m}{v}}\\ =-0.014$

The negative sign signifies with an increase in temperature density will decrease.

Power of the drilling machine, P = 10 kW

Time. t = 2.5 min

Total energy dissipated E is

$\\E=Pt\\ E=10\times 10^{3}\times 2.5\times 60\\ E=1.5\times 10^{6}J$

Thermal energy absorbed by aluminium block is

$\\\Delta Q=\frac{E}{2}\\ \Delta Q=\frac{1.5\times 10^{6}}{2}\\ \Delta Q=7.5\times 10^{5}J$

Mass of the aluminium block, m = 8.0 kg

Specific heat of aluminium, c = 0.91 J g-1 K-1

Let rise in temperature be $\Delta T$

$\\mc\Delta T=\Delta Q\\ \Delta T=\frac{\Delta Q}{mc}\\ \Delta T=\frac{7.5\times 10^{5}}{0.91\times 8\times 10^{3}}\\ \Delta T=103.02\ ^{o}C$

Mass of copper block m = 2.5 kg

Initial Temperature of the copper block, T1 = 500 oC

Final Temperature of Copper block, T2 = 0 oC

Specific heat of copper, c = 0.39 J g-1 K-1

Thermal Energy released by the copper block is $\Delta Q$

$\\\Delta Q=mc\Delta T\\ \Delta Q=2.5\times 10^{3}\times 0.39\times 500\\ \Delta Q=487500\ J$

Latent heat of fusion of water, L = 335 j g-1

Amount of ice that can melt is

$\\w=\frac{\Delta Q}{L}\\ w=\frac{487500}{335}\\ w=1455.22g$

1.455 kg of ice can melt using the heat released by the copper block.

Let the specific heat of the metal be C.

Mass of metal block m = 200 g

Initial Temperature of metal block = 150 oC

Final Temperature of metal block = 40 oC

The heat released by the block is

$\\\Delta Q=mc\Delta T\\ \Delta Q=200\times c\times (150-40)\\ \Delta Q=22000c$

Initial Temperature of the calorimeter and water = 27 oC

Final Temperature of the calorimeter and water = 40 oC

Amount of water = 150 cm

Mass of water = 150 g

Water equivalent of calorimeter = 25 g

Specific heat of water = 4.186 J g-1 K-1

Heat absorbed by the Calorimeter and water is $\\\Delta Q'$

$\\\Delta Q'=(150+25)\times 4.186\times (40-27)\\ \Delta Q'=9523.15J$

The heat absorbed by the Calorimeter and water is equal to the heat released by the block

$\\\Delta Q=\Delta Q'\\ 22000c=9523.15\\ c=0.433\ J\ g^{-1}\ K^{-1}$

The above value would be lesser than the actual value since some heat must have been lost to the surroundings as well which we haven't accounted for.

## Thermal Properties of Matter Excercise:

### Question:

Gas                                    Molar specific heat (Cv )

$\inline (cal\; mol^{-1}K^{-1})$

Hydrogen                                      4.87

Nitrogen                                        4.97

Oxygen                                         5.02

Nitric oxide                                    4.99

Carbon monoxide                          5.01

Chlorine                                         6.17

The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is  $\inline 2.92\; cal/mol\; K.$  Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

Monoatomic gases have only translational degree of freedom but diatomic gases have rotational degrees of freedom as well. The temperature increases with increase in the spontaneity of motion in all degrees. Therefore to increase the temperature of diatomic gases more energy is required than that required to increase the temperature of monoatomic gases by the same value owing to higher degrees of freedom in diatomic gases.

If we only consider rotational modes of freedom the molar specific heat of the diatomic gases would be given as

$\\c=\frac{fR}{2}\\ c=\frac{5}{2}\times 1.92\\ c=4.95\ cal\ mol^{-1}\ K^{-1}$

The number of degrees of freedom = 5 (3 translational and 2 rotational)

The values given in the table are more or less in accordance with the above calculated one. The larger deviation from the calculated value in the case of chlorine is because of the presence of vibrational motion as well.

Initial Temperature of the boy = 101 oF

Final Temperature of the boy = 98 oF

Change in Temperature is

$\\\Delta T=3\ ^{o}F\\ \Delta T=3\times \frac{5}{9}\\ \Delta T=1.67\ ^{o}C$

Mass of the child is m = 30 kg

Specific heat of human body = 1000 cal kg-1 oC-1

Heat released is $\\\Delta Q$

$\\\Delta Q=mc\Delta T\\\Delta Q=30\times 1000\times 1.67\\ \Delta Q=50000\ cal$

Latent heat of evaporation of water = 580 cal g-1

The amount of heat lost by the body of the boy has been absorbed by water.

Let the mass of water which has evaporated be m'

$\\\Delta Q=m'L\\ m'=\frac{Q}{L}\\ m'=\frac{50000}{580}\\ m'=86.2\ g$
Time in which the water has evaporated, t = 20 min.

Rate of evaporation is m'/t

$\\\frac{m'}{t}=\frac{86.2}{20}\\ \frac{m'}{t}=4.31\ g\ min^{-1}$

Side of the box s = 30 cm

Area available for conduction A

A = 6s2

A=6(30)2

A=5400 cm2 = 0.54 m2

Temperature difference = 45 oC

Co-efficient of thermal conductivity of thermacole is k = 0.01 J s-1 m-1 K-1

Width of the box is d = 5 cm

Heat absorbed by the box in 6 hours is $\Delta Q$

$\\\Delta Q=\frac{kA\Delta T}{l}\\ \Delta Q=\frac{0.01\times 0.54\times 45\times 6\times 60\times 60}{0.05}\\ \Delta Q=104976\ J$

The heat of fusion of water is $L=335\times 10^{3}\ J\ kg^{-1}$

Amount of ice which has melted is m'

$\\m'=\frac{\Delta Q}{L}\\ m'=\frac{104976}{335\times 10^{3}}\\ m'=0.313\ kg$

Amount of ice left after 6 hours = 4 - 0.313 = 3.687 kg

The rate at which water boils, R = 6.0 kg min-1

The heat of vaporisation of water, $L=2256\times 10^{3}\ J\ kg^{-1}$

The rate at which heat enters the boiler

$\\\frac{dQ}{dT}=RL\\ \frac{dQ}{dT}=\frac{6\times 2256\times 10^{3}}{60}\\ \frac{dQ}{dT}=2.256\times 10^{5}Js^{-1}$

The base area of the boiler, A = 0.15 m2

Thickness, l = 1.0 cm

Thermal conductivity of brass$\inline =109\; j\; s^{-1}m^{-1}K^{-1};$

The temperature inside the boiler = Boiling point of water = 100 oC

Let the temperature of the flame in contact with the boiler be T

Amount of heat flowing into the boiler is

$\\\frac{dQ}{dt}=\frac{KA\Delta T}{l}\\ 2.256\times 10^{5}=\frac{109\times 0.15\times (T-100)}{1\times 10^{-2}}\\ T-100=137.98\\ T=237.98\ ^{o}C$

The temperature of the flame in contact with the boiler is 237.98 oC

Q. 11.19 (a) Explain why :

(a) a body with large reflectivity is a poor emitter

A body with a large reflectivity is a poor absorber. As we know a body which is a poor absorber will as well be a poor emitter. Therefore a body with large reflectivity is a poor emitter.

Q .11.19 (b) Explain why :

(b) a brass tumbler feels much colder than a wooden tray on a chilly day

Brass is a good conductor of heat. Therefore once someone touches brass heat from their body flows into it and it feels cold, in case of a wooden tray, no such conduction of heat from the body takes place as wood is a very poor conductor of heat.

Q. 11.19 (c) Explain why :

(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open but gives a correct value for the temperature when the same piece is in the furnace

An optical pyrometer relates the brightness of a glowing body with its temperature. In the open because of other sources of light the sensor in the optical pyrometer does not detect the true brightness of a red hot piece of iron and thus does not predict its temperature correctly whereas in the furnace the piece of iron is the only source of light and the sensor detects its brightness correctly thus giving the correct value of the temperature.

Q. 11.19 (d) Explain why :

(d) the earth without its atmosphere would be inhospitably cold

The sun rays contain infrared radiations. These are reflected back by the lower part of the atmosphere after being reflected by the surface of the earth and are trapped inside the atmosphere thus maintaining the Earth's temperature at a hospitable level. Without these rays being trapped the temperature of the earth will go down severely and thus the Earth without its atmosphere would be inhospitably cold.

Q. 11.19 (e) Explain why :

(e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water

Heating systems based on the circulation of steam are more efficient in warming a building than those based on the circulation of hot water because the same amount of steam at 100 oC contains more energy available for heat dissipation than the same amount of water at 100 oC in the form of latent heat of vaporization.

Let a body initially be at temperature T1

Let its final Temperature be T2

Let the surrounding temperature be T0

Let the temperature change in time t.

According to Newton's Law of cooling

$\\-\frac{dT}{dt}=K(T-T_{0})\\ \frac{dT}{T-T_{0}}=-Kdt\\ \int_{T_{1}}^{T_{2}}\frac{dT}{T-T_{0}}=-K\int_{0}^{t}dt\\ \left [ ln(T-T_{0}) \right ]_{T_{1}}^{T_{2}}=-K[t]_{0}^{t}\\ ln\left ( \frac{T_{2}-T_{0}}{T_{1}-T_{0}} \right )=-Kt$

where K is a constant.

We have been given that the body cools from 80 oC to 50 oC in 5 minutes when the surrounding temperature is 20 oC.

T2 = 50 oC

T1 = 80 oC

T0 = 20 oC

t = 5 min = 300 s.

$\\ln\left ( \frac{50-20}{80-20} \right )=-K\times 300\\ K=\frac{ln(2)}{300}\\$

For T1 = 60 oC and T2 = 30 oC we have

$\\ln\left ( \frac{30-20}{60-20} \right )=-\frac{ln2}{300}t\\ t=\frac{ln(4)\times 300}{ln(2)} \\t=\frac{2ln(2)\times 300}{ln(2)}\\ t=600\ s\\ t= 10\ min$

The body will take 10 minutes to cool from 60 oC to 30 oC at the surrounding temperature of 20 oC.

NCERT solutions for class 11 physics chapter 11 thermal properties of matter additional exercise

(a) At what temperature and pressure can the solid, liquid and vapour phases of  $\inline CO_{2}$  co-exist in equilibrium?

At the triple point temperature of -56.6 oC and pressure 5.11 atm the solid, liquid and vapour phases of  $\inline CO_{2}$  co-exist in equilibrium.

(b) What is the effect of decrease of pressure on the fusion and boiling point of $CO_{2}$ ?

Both fusion and boiling point of CO2 decrease with decrease in pressure. This we can see from the solid lines in the P-T phase diagram of CO2.

(c) What are the critical temperature and pressure for  $\inline CO_{2}$ ? What is their significance?

The critical temperature and pressure for CO2 are 31.1 oC and 73.0 atm respectively. If the temperature exceeds this critical value of temperature CO2 would not liquefy no matter how high the pressure is.

(d) Is $\inline CO_{2}$  solid, liquid or gas at

(a) $\inline -70^{\circ}C$ under $\inline 1\; atm,$

CO2 is vapour at -70 oC under 1 atm pressure as the point corresponding to this condition lies in vapour region in the given P-T phase diagram of carbon dioxide.

(d) Is $\inline C0_{2}$ solid, liquid or gas at

(b) $\inline -60^{\circ}C$ under $\inline 10 \; atm,$

CO2 is solid at -60 oC under 10 atm pressure as the point corresponding to this condition lies in the solid region in the given P-T phase diagram of carbon dioxide.

(d) Is $\inline CO_{2}$ solid, liquid or gas at

(c) $\inline 15^{\circ}C$ under $\inline 56\; atm\; ?$

CO2 is liquid at 15 oC under 56 atm pressure as the point corresponding to this condition lies in the liquid region in the given P-T phase diagram of carbon dioxide.

## Q. 11.22 (a) Answer the following questions based on the P – T phase diagram of $\inline CO_{2}$ :

(a) $\inline CO_{2}$  at 1 atm pressure and temperature $\inline -60^{\circ}C$  is compressed isothermally. Does it go through a liquid phase?

The temperature -60 oC lies to the left of the triple point of water i.e. in the region of solid and vapour phases. Once we start compressing CO2 at this temperature starting from 1 atm pressure it will directly convert into solid without going through the liquid phase.

## Q. 11.22 (b) Answer the following questions based on the P – T phase diagram of $CO_{2}$ :

(b) What happens when $\inline CO_{2}$  at $\inline 4$ atm pressure is cooled from room temperature at constant pressure?

At room temperature (27 oC) and 4 atm pressure CO2 exits in the vapour phase. The pressure 4 atm is less than the pressure at the triple point and therefore points corresponding to all temperatures and this pressure lie in the solid and vapour region. Once we start compressing CO2 from room temperature at this constant pressure CO2 turns from vapour to solid directly without going through the liquid phase.

(c) Describe qualitatively the changes in a given mass of solid $\inline CO_{2}$  at $\inline 10\; atm$ pressure and temperature $\inline -65^{\circ}C$ as it is heated up to room temperature at constant pressure.

At -65 oC under 10 atm pressure CO2 is in the solid phase. At room temperature (27 oC) under 10 atm pressure CO2 is in the vapour phase. At 10 atm pressure, CO2 can exist in all three phases depending upon the temperature. Therefore as CO2 is heated from -65 oC to room temperature at a constant pressure of 10 atm it goes from the solid phase to liquid phase and then ultimately it goes into the vapour phase.

(d) $\inline CO_{2}$ is heated to a temperature $\inline 70^{\circ}C$ and compressed isothermally. What changes in its properties do you expect to observe?

70 oC is above the critical temperature of CO2. Once CO2 is isothermally compressed at this temperature it would not liquefy irrespective of how high the pressure is but at very high pressures CO2 will not behave as an ideal gas.

## NCERT solutions for class 11 physics chapter wise

 Chapter 1 NCERT solutions for class 11 physics chapter 1 Physical world Chapter 2 Solutions of NCERT for class 11 physics chapter 2 Units and Measurement Chapter 3 CBSE NCERT solutions for class 11 physics chapter 3 physics Motion in a straight line Chapter 4 NCERT solutions for class 11 physics chapter 4 Motion in a Plane Chapter 5 Solutions of NCERT for class 11 physics chapter 5 Laws of Motion Chapter 6 CBSE NCERT solutions for class 11 physics chapter 6 Work, Energy and Power Chapter 7 NCERT solutions for class 11 physics chapter 7 System of Particles and Rotational motion Chapter 8 Solutions of NCERT for class 11 physics chapter 8 Gravitation Chapter 9 CBSE NCERT solutions for class 11 physics chapter 9 Mechanical Properties of Solids Chapter 10 NCERT solutions for class 11 physics chapter 10 Mechanical Properties of Fluids Chapter 11 NCERT solutions for class 11 physics chapter 11 Thermal Properties of Matter Chapter 12 CBSE NCERT solutions for class 11 physics chapter 12 Thermodynamics Chapter 13 NCERT solutions for class 11 physics chapter 13 Kinetic Theory Chapter 14 Solutions of NCERT for class 11 physics chapter 14 Oscillations Chapter 15 CBSE NCERT solutions for class 11 physics chapter 15 Waves

## Subject wise solutions for NCERT class 11

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## Importance of NCERT solutions for class 11 physics chapter 11 thermal properties of matter:

Thermal properties of matter is an important chapter for competitive exams. Questions based on calorimetry heat transfer and thermal expansions are frequently asked in competitive exams like NEET and JEE Mains exam. The NCERT solutions for class 11 physics chapter 11 thermal properties of matter are helpful to understand these concepts and their application in numerical problems.