Your Progress — Chapter 3: Thermal Physics
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MCQI. Multiple Choice Questions1 mark each
Q.1 The value of universal gas constant
✓ Answer: (A)
Q.2 If a substance is heated or cooled, the change in mass of that substance is
✓ Answer: (A)
Q.3 If a substance is heated or cooled, the linear expansion occurs along the axis of
✓ Answer: (A)
Q.4 Temperature is the average ___________ of the molecules of a substance
✓ Answer: (A)
Q.5 In the Given diagram, the possible direction of heat energy transformation is
✓ Answer: (A)
FillII. Fill in the Blanks1 mark each
| # | Statement (Answer in bold) |
|---|---|
| 1 | The value of Avogadro number 6.023 x 1023 / mol (or) mol-1 |
| 2 | The temperature and heat are Scalar quantities |
| 3 | One calorie is the amount of heat energy required to raise the temperature of 1 gm of water through 1^\circ C. |
| 4 | According to Boyle’s law, the shape of the graph between pressure and reciprocal of volume is straight line |
T/FIII. True or False1 mark each
| # | Statement | Answer | Correction (if False) |
|---|---|---|---|
| 1 | For a given heat in liquid, the apparent expansion is more than that of real expansion. | False | The real expansion is more (or) less than that of apparent expansion. |
| 2 | Thermal energy always flows from a system at higher temperature to a system at lower temperature. | True | — |
| 3 | According to Charles’s law, at constant pressure, the temperature is inversely proportional to volume. | False | Volume is directly proportional to temperature at constant pressure. |
MatchIV. Match the Following1 mark each
| Column A | Column B |
|---|---|
| Linear expansion | Change in length |
| Superficial expansion | change in area |
| Cubical expansion | change in volume |
| Heat transformation | hot body to cold body |
| Boltzmann constant | 1.381 x 10-23 JK-1 |
A&RV. Assertion & Reasoning2 marks each
Q.1
Assertion: If one end of the rod is heated,other end also is heated.
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✓ Answer
Reason: Heat always flows from a region of lower temperature to higher temperature of the rod.
Answer: c. Assertion is true but the reason is false.
Explanation: The assertion is true because thermal conduction ensures heat travels to the other end of a metal rod. The reason is completely false because heat naturally flows from a higher temperature to a lower temperature.
Answer: c. Assertion is true but the reason is false.
Explanation: The assertion is true because thermal conduction ensures heat travels to the other end of a metal rod. The reason is completely false because heat naturally flows from a higher temperature to a lower temperature.
Q.2
Assertion: Gas is highly compressible than solid and liquid
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✓ Answer
Reason: Interatomic or intermolecular distance in the gas is comparably high.
Answer: a) Both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: Gases are highly compressible precisely because the spaces (intermolecular distances) between their particles are huge compared to solids and liquids, leaving plenty of room to push them closer together.
Answer: a) Both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: Gases are highly compressible precisely because the spaces (intermolecular distances) between their particles are huge compared to solids and liquids, leaving plenty of room to push them closer together.
ShortVI. Short Answer Questions2 marks each
Q.1
Define one calorie.
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✓ Answer
One calorie is the amount of heat required to rise the temperature of 1 gram of water through 1^\circ C.
Q.2
Distinguish between linear, arial and superficial expansion.
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✓ Answer
Liner Expansion
Q.1
Also called longitudinal expansion
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✓ Answer
$2. \alpha L = \Delta L/L0\Delta T$
arial Expansion and superficial expansion
arial Expansion and superficial expansion
Q.1
Also called as superficial expansion
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✓ Answer
$2. \alpha A = \Delta A/A0\Delta T$
Q.3
What is co-efficient of cubical expansion?
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✓ Answer
The ratio of increase in volume of the body per degree rise in temperature to its unit volume is called as co-efficient of cubical expansion. It's unit is K-1.
Q.4
State Boyle’s law
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✓ Answer
When the temperature is kept constant, the volume of a fixed mass of gas is inversely proportional to its pressure. P \propto 1/V
$PV = constant$
$PV = constant$
Q.5
State-the law of volume
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✓ Answer
According to this law, when the pressure of gas is kept constant, the volume of a gas is directly proportional to the temperature of the gas.
$V \propto T or V/T = constant$
$V \propto T or V/T = constant$
Q.6
Distinguish between ideal gas and real gas.
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✓ Answer
Ideal gas
Q.1
If the atoms or molecules of a gas do not interact with each other, then the gas is said to be an ideal gas or a perfect gas.
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✓ Answer
Refer to textbook.
Q.2
Ideal gas does not have volume.
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✓ Answer
Real gas
Q.1
If the molecules or atoms of a . gas interact with each other with a definite amount of intermolecular or inter atomic force of attraction, then the gas is said to be a real gas.
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✓ Answer
Refer to textbook.
Q.2
Real gas has volume.
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✓ Answer
Refer to textbook.
Q.7
What is co-efficient of real expansion?
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✓ Answer
The ratio of true rise in the volume of the liquid to unit original volume when the temperature rises by one kelvin is called as “Coefficient of real expansion”. Its SI unit is per K or K-1.
Q.8
What is co-efficient of apparant expansion?
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✓ Answer
Co-efficient of apparent expansion is defined as the ratio of the apparent rise in the volume of the liquid per degree rise in temperature to its unit volume. The SI unit of coefficient of apparent expansion is K-1.
NumericalVII. Numerical Problems3 marks each
Q.1
Find the final temperature of a copper rod. Whose area of cross section changes from 10 m2 to 11 m2 due to heating. The copper rod is initially kept at 90 K. (Coefficient of superficial expansion is 0.0021 /K)
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✓ Answer
Given
$Area of copper rod, A = 10 m2$
$After expansion, A2 = 11 m2$
$Initial temperature T1 = 90 K$
$Coefficient of superficial expansion = 0.0021 / K$
$To find: Final temperature T2 = ?$
Solution
$\Delta A/A = \alpha \Delta T$
$(A2-A1)/A = \alpha \Delta (T2-T1)$
$11-10 / 10 = 0.0021(T2 - 90)$
$1 / [ 10 x 0.0021 ] = (T2 - 90)$
$T2 = (1/0.021) + 90 = 47.6 + 90$
$Final Temperature, T2 = 137.6 K$
$Area of copper rod, A = 10 m2$
$After expansion, A2 = 11 m2$
$Initial temperature T1 = 90 K$
$Coefficient of superficial expansion = 0.0021 / K$
$To find: Final temperature T2 = ?$
Solution
$\Delta A/A = \alpha \Delta T$
$(A2-A1)/A = \alpha \Delta (T2-T1)$
$11-10 / 10 = 0.0021(T2 - 90)$
$1 / [ 10 x 0.0021 ] = (T2 - 90)$
$T2 = (1/0.021) + 90 = 47.6 + 90$
$Final Temperature, T2 = 137.6 K$
Q.2
Calculate the coefficient of cubical expansion of a zinc bar. Whose volume is increased 0.25 m3 from 0.3 m3 due to the change in its temperature of 50 K.
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✓ Answer
Given
$Volume of the zinc bar V = 0.3 m3$
$Change in volume \Delta V = 0.25 m3$
$Change in temperature \Delta T = 50 K$
$To find : Coefficient of cubic expansion = ?$
Solution
$\Delta V/V = \alpha V\Delta T$
$\alpha V = \Delta V / V. \Delta T$
$\alpha V = 0.25 / (0.3 x 50) = 0.25 / 15 = 0.0166 K-1$
Coefficient of cubical expansion,
$\alpha V = 0.0166 K-1.$
$Volume of the zinc bar V = 0.3 m3$
$Change in volume \Delta V = 0.25 m3$
$Change in temperature \Delta T = 50 K$
$To find : Coefficient of cubic expansion = ?$
Solution
$\Delta V/V = \alpha V\Delta T$
$\alpha V = \Delta V / V. \Delta T$
$\alpha V = 0.25 / (0.3 x 50) = 0.25 / 15 = 0.0166 K-1$
Coefficient of cubical expansion,
$\alpha V = 0.0166 K-1.$
LongVIII. Long Answer Questions5 marks each
Q.1
Derive the ideal gas equation.
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✓ Answer
(i) The ideal gas equation is an equation which relates all the properties of an ideal gas such as pressure (P), volume (V), temperature (T) and the amount of the gas.
(ii) An ideal gas obeys Boyle’s law and Charles’s law and Avogadro’s law.
(iii) According to Boyle’s law,
$PV = constant ... (1)$
(iv) According to Charles’s law,
$V/T = constant ... (2)$
(v) According to Avogadro’s law,
$V/n = constant ... (3)$
(vi) After combining equations (1), (2) and (3), we can write the following equation.
$PV/nT = constant ... (4)$
The above relation is called the combined gas law.
(vii) If we consider a gas which contains \mu moles of the gas, the number of atoms contained will be equal to \mu times the Avogadro number, NA. i.e., n = \mu NA.
$(viii) Using this value, equation (4) can be written as PV/ \mu NAT = constant$
(ix) The value of constant in the above equation is taken to be kB, which is called as Boltzmann’s constant.
$Its value is 1.381 \times 10-23 JK-1.$
(x) Hence, we have the following equation:
$PV/ \mu NAT = kB$
$PV = \mu NA kB T$
$(xi) Here, \mu NA kB = R, which is termed as universal gas constant whose value is 8.31$
J mol-1K-1.
$PV = RT ... (5)$
This is called ideal gas equation.
(ii) An ideal gas obeys Boyle’s law and Charles’s law and Avogadro’s law.
(iii) According to Boyle’s law,
$PV = constant ... (1)$
(iv) According to Charles’s law,
$V/T = constant ... (2)$
(v) According to Avogadro’s law,
$V/n = constant ... (3)$
(vi) After combining equations (1), (2) and (3), we can write the following equation.
$PV/nT = constant ... (4)$
The above relation is called the combined gas law.
(vii) If we consider a gas which contains \mu moles of the gas, the number of atoms contained will be equal to \mu times the Avogadro number, NA. i.e., n = \mu NA.
$(viii) Using this value, equation (4) can be written as PV/ \mu NAT = constant$
(ix) The value of constant in the above equation is taken to be kB, which is called as Boltzmann’s constant.
$Its value is 1.381 \times 10-23 JK-1.$
(x) Hence, we have the following equation:
$PV/ \mu NAT = kB$
$PV = \mu NA kB T$
$(xi) Here, \mu NA kB = R, which is termed as universal gas constant whose value is 8.31$
J mol-1K-1.
$PV = RT ... (5)$
This is called ideal gas equation.
Q.2
Explain the experiment of measuring the real and apparent expansion of a liquid with a neat diagram.
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✓ Answer
(i) The liquid whose real and apparent expansion is to be determined is poured in a container up to a level. Mark this level as L1.
(ii) Now, heat the container and the liquid using a burner. Initially, the container receives the thermal energy and it expands.
(iii) As a result, the volume of the liquid appears to have reduced. Mark this reduced level of liquid as L2.
(iv) On further heating, the thermal energy is supplied to the liquid resulting in the expansion of the liquid. Hence, the level of liquid rises to L3.
(v) Now, the difference between the levels L1 and L3 is called as apparent expansion, and the difference between the levels L2 and L3 is called real expansion.
(vi) The real expansion is always more than the apparent expansion.
$(vii) Real expansion = L3 - L2$
$Apparent expansion = L3 - L1$
(ii) Now, heat the container and the liquid using a burner. Initially, the container receives the thermal energy and it expands.
(iii) As a result, the volume of the liquid appears to have reduced. Mark this reduced level of liquid as L2.
(iv) On further heating, the thermal energy is supplied to the liquid resulting in the expansion of the liquid. Hence, the level of liquid rises to L3.
(v) Now, the difference between the levels L1 and L3 is called as apparent expansion, and the difference between the levels L2 and L3 is called real expansion.
(vi) The real expansion is always more than the apparent expansion.
$(vii) Real expansion = L3 - L2$
$Apparent expansion = L3 - L1$
HOTIX. Higher Order Thinking3 marks each
Refer to textbook for answers.
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