NCERT Solution (Class 11) > Physics Chapter 11 Thermal Properties Of Matter
NCERT Solution Class 11 Physics Chapter 11 Thermal Properties Of Matter: National Council of Educational Research and Training Class 11 Physics Chapter 11 Solution – Thermal Properties Of Matter. Free PDF Download facility available at our website.
Board |
NCERT |
Class |
11 |
Subject |
Physics |
Chapter |
11 |
Chapter Name |
Thermal Properties Of Matter |
Topic |
Exercise Solution |
Thermal Properties Of Matter Chapter all Questions and Numericals Solution
Chapter11- Thermal Properties Of Matter
11.1) The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
ANSWER-
Given temperatures are expressed in the Kelvin scale. So we know the relation between Celsius and Kelvin scale
TC = TK – 273.15 ……………1
And relation between Celsius and Fahrenheit scales is
TF = 9/5 x Tc + 32 ……………..2
11.2)
11.3)
11.4) Answer the following:
(a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?
(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale?
(c) The absolute temperature (Kelvin scale) T is related to the temperature tc on the Celsius scale by
tc = T – 273.15Why do we have 273.15 in this relation, and not 273.16 ?
(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?
ANSWER-
(a) The triple point of water is fixed point because it has a unique value which is 273.16 K At particular values of volume and pressure. On the other hand The melting point of ice and boiling point of water depend on pressure and temperature hence they do not have unique values and can not be used as standard fixed points.
(b) The absolute zero is the other fixed point on the Kelvin absolute scale.
(c) The temperature 273.16 K is the triple point of water while the temperature 273.15 K is the melting point of water. Hence In this relation tc = T – 273.15we have 273.15 to represents equivalent of o Celsius of melting point of ice and not 273.16 .
(d) Let TF be the temperature on Fahrenheit scale and TK be the temperature on absolute scale. Then relation is given by
11.5) 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 | 1.250 × 105 Pa | 0.200 × 105 Pa |
Normal melting point of sulphur | 1.797 × 105 Pa | 0.287 × 105 Pa |
(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B ?
(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?
ANSWER-
(a)At triple point of water (T), pressure in thermometer A, pA = 1.250 × 105 Pa and pressure in thermometer B, pB = 0.2 × 105 Pa
At melting point of sulphur (T1), pressure in thermometer A, p1 = 1.797× 105 Pa and pressure in thermometer B, p2 = 0.287 × 105 Pa
For thermometer A ;
Therefore, the absolute temperature of the normal melting point of sulphur as Read by thermometer A is 392.69 K. and read by thermometer B is 391.98 K.
(b)The oxygen and hydrogen gas present in thermometers A and B are not perfect ideal gases. Hence, there is a slight difference between the readings of thermometers A and B. To reduce the discrepancy between the two readings, the experiment should be carried under low pressure conditions as gases behave as perfect ideal gases at low pressure and high temperature.
11.6)
11.7)
11.8)
11.9)
11.10)
11.11)
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11.13)
11.15) Given below are observations on molar specific heats at room temperature of some common gases.
Gas |
Molar specific heat (Cv) (cal mo1–1 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 is2.92 cal/mol K. Explain this difference. What can you infer from the some what larger (than the rest) value for chlorine?
ANSWER-
Monoatomic gases have only translational degree of freedom on the other hand diatomic gases mentioned in the above table have translational degree of freedom along with rotational degree of freedom and some may have vibrational degree of freedom hence heat must be supplied to increase average energy of all the modes (translational, rotational or vibrational) hence the measured molar specific heats of these gases are markedly different from those for monatomic gases.
11.16)
11.17)
11.18)
11.19) Explain why:
(a) a body with large reflectivity is a poor emitter
(b) a brass tumbler feels much colder than a wooden tray on a chilly day
(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
(d) The earth without its atmosphere would be inhospitably cold
(e) Heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.
ANSWER-
(a) A body with a large reflectivity is a poor absorber of light radiations. A poor absorber will absorb less radiations hence emit less. Hence, a body with a large reflectivity is a poor emitter.
(b) Brass is a good conductor of heat. When one touches a brass tumbler, heat is transferred from the body to the brass tumbler easily. Hence, the temperature of the body reduces immediately and one feels cooler.
Wood is a poor conductor of heat. When one touches a wooden tray, very little heat is conducted from the body to the wooden tray. Hence, there is small temperature decrement of the body hence a brass tumbler feels colder than a wooden tray on a chilly day.
(c) For black body; radiation energy (E) is given by :
E α (T4 – T04)
Where;
T = Temperature of pyrometer.
T0 = Temperature of open space.
Hence, an increase in the temperature of open space reduces the radiation energy hence an optical pyrometer gives too low value for the temperature of a red hotiron piece in the open.
(d) In the absence of atmospheric gases, All the heat coming from sun would be radiated back from earth’s surface and no heat will get trapped hence the earth without its atmosphere would be inhospitably cold.
(e) When hot water heated further it transforms into steam. Steam contains extra heat called latent heat. Hence A heating system based on the circulation of steam is more efficient in warming a building than that based on the circulation of hot water.
11.20)
Additional Exercises
11.21) Answer the following questions based on the P-T phase diagram of carbon dioxide:
(a) At what temperature and pressure can the solid, liquid and vapour phases of CO2 co-exist in equilibrium?
(b) What is the effect of decrease of pressure on the fusion and boiling point of CO2?
(c) What are the critical temperature and pressure for CO2 ? What is their significance?
(d) Is CO2 solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60 °C under 10 atm?(c) 15°C under 56 atm ?
ANSWER-
P-T phase diagram of carbon dioxide is given below:
(a) From P-T phase diagram of carbon dioxide point O is triple point of co2 where all the tree phases coexist. From diagram point O is corresponding to pressure of 5.11 atm and temperature of -56.6 0c.
(b)The fusion and boiling point of CO2 decreases with decrease in pressure.
(c)The critical temperature of gas is the temperature at and above which vapor of the gas can not be liquefied independent of how much pressure is applied. Hence from diagram The critical temperature and critical pressure of co2 are 31.10c and 73 atm respectively. It signifies that co2 will not liquefy above 31.10c no matter whatever may be the pressure.
(d) From P-T phase diagram of CO2;
CO2 is gaseous at –70°C, under 1 atm pressure.
CO2 is solid at –60°C, under 10 atm pressure.
CO2 is liquid at 15°C, under 56 atm pressure.
11.22) Answer the following questions based on the P – T phase diagram of CO2:
(a) CO2 at 1 atm pressure and temperature – 60 °C is compressed isothermally. Does it go through a liquid phase?
(b) What happens when CO2 at 4 atm pressure is cooled from room temperature at constant pressure?
(c) Describe qualitatively the changes in a given mass of solid CO2 at 10 atm pressure and temperature –65 °C as it is heated up to room temperature at constant pressure.
(d) CO2 is heated to a temperature 70 °C and compressed isothermally. What changes in its properties do you expect to observe?
ANSWER-
(a) FROM P-T diagramas it lies in the region of vaporous and solid phases: It condenses to solid directly from vapour state.
(b) At 4 atm pressure, CO2 lies below triple point O. when it cools down at constant pressure it directly converted into solid states as it lies in the region of vaporous and solid phases.
(c) When the temperature of a mass of solid CO2 at 10 atm pressure and at –65°C is increased at constant pressure, it converts into liquid state and then vapour state as temperature increases.
(d) If CO2 is heated to 70°C and compressed isothermally, as 70°C is higher than the critical temperature of CO2 (-56.60c). hence it will remain in the vapour state.
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