Contents
- 1 Introduction: Expansion of Liquids
- 1.1 Apparent And Real Expansion of Liquids
- 1.2 Coefficients of Apparent and Real Expansions of Liquids
- 1.3 Coefficient of apparent expansion of a liquid:
- 1.4 Coefficient of real expansion of a liquid:
- 1.5 Coefficient of real expansion
- 1.6 Relation between the Coefficients of Apparent and Real Expansions of Liquids
- 1.7 Relation between Density and Coefficient of Real Expansion of Liquids
- 1.8 Numerical Examples
Some of the most important Physics Topics include energy, motion, and force.
Introduction: Expansion of Liquids
Liquids expand with the rise in temperature just like solids. The solids have definite shapes. Hence, in case of solids three types of expansion (linear, surface and volume) are significant. But liquids have no definite shape. Thus unlike solids, liquid expansion with change in temperature is studied only in terms of the change in volume. Other characteristic features in liquid expansion are
- For the same rise in temperature, thermal expansion of a liquid is about ten times that of a solid of the same volume.
- For the same rise in temperature, different liquids of equal volume expand differently.
- The rate of thermal expansion differs a little for the different ranges of temperature change.
Example: Expansion of water is different for temperature ranges 10°C to 11°C and 93°C to 94°C. Moreover, water contracts in volume when temperature increases from 0°C to 4°C.
Apparent And Real Expansion of Liquids
To heat a liquid, it has to be kept in a container. When heat is applied, the container also expands along with the liquid. As the liquid expands more than the container for the same change in temperature, expansion of the container is sometimes neglected and only the expansion of the liquid is recorded. Hence, the recorded expansion of the liquid, ignoring the expansion of the container, is less than the actual expansion of the liquid.
The expansion of a liquid, ignoring the expansion of the container, is called the apparent expansion of the liquid. The sum of the apparent expansion of the liquid and the expansion of the container, is called the real expansion of the liquid.
Experiment: Let, surface of the liquid in a flask rests at mark O [Fig.]. If the flask is heated from outside, at first the flask expands and the surface of the liquid comes down at mark A. Thus the length OA represents expansion of the flask. Then the supplied heat reaches the liquid and surface of the liquid rises up to mark B. Therefore, volume of AB is the real expansion of the liquid.
On the other hand, if the first change of volume is unnoticed, it seems like the surface rises from initial position O to final position B. So, volume of OB is the apparent expansion of the liquid.
∴ For a liquid, real expansion = apparent expansion of the liquid + expansion of container
Coefficients of Apparent and Real Expansions of Liquids
Since liquid expansions are of two types, two separate coef-ficients of expansion are to be considered:
- Coefficient of apparent expansion and
- coefficient of real expansion.
Coefficient of apparent expansion of a liquid:
Definition: The apparent expansion of unit volume of a liquid for a temperature rise of 1° is called the coefficient of apparent expansion (γ’) of the liquid.
Expression for γ’: Let the volume of a certain amount of liquid be V1 at temperature t1, and its apparent volume be V2‘ at temperature t2.
∴ For a rise in temperature of (t2 – t1), apparent expansion of the liquid of volume V1 = (V2‘ – V1).
∴ For a unit rise in temperature, apparent expansion of the liquid of volume V1 = \(1\)
For a unit rise in temperature, the apparent expansion per unit volume = \(\frac{V_2{ }^{\prime}-V_1}{V_1\left(t_2-t_1\right)}\)
By definition,
γ’ = \(\frac{V_2{ }^{\prime}-V_1}{V_1\left(t_2-t_1\right)}\)
\(=\frac{\text { apparent expansion }}{\text { initial volume } \times \text { rise in temperature }}\) ……… (1)
From (1), we get, V2’ = V1 {1 + γ’\t2 – t1} ……. (2)
It is important to note that, the coefficient of apparent expansion of a liquid is not an intrinsic property of the liquid. It depends on the material of the container. Hence, a liquid may have different values of y’ when heated in containers of different materials.
Coefficient of real expansion of a liquid:
Definition: The actual or real increase of unit volume of a liquid for a temperature rise of 1° is called the coefficient of real expansion (γ) of the liquid.
Expression for γ : Let the volume of a fixed amount of a liquid at a temperature t1 be V1, and at a temperature t2 be V2.
Hence, volume V1 increases by (V2 – V1) for a rise (t2 – t1) in temperature.
∴ By definition,
γ = \(\frac{V_2-V_1}{V_1\left(t_2-t_1\right)}\)
\(=\frac{\text { real expansion }}{\text { initial volume } \times \text { rise in temperature }}\) …….. (3)
∴ V2 = V1{1 + γ(t2 – t1)} ……. (4)
The coefficient of real expansion is an intrinsic property of the liquid and does not depend on the material of the container.
1. It is clear from equations (1) and (3) that the values of γ and γ’ are independent of the unit of volume but depend on scale of temperature used.
For example, the coefficient of real expansion of mercury in the Celsius and the Fahrenheit scales are 18.18 × 10-5°C-1 and 10.1 × 10-5°F-1 respectively. The coefficient of volume expansion of a liquid is the same in the Celsius and the Kelvin scales but in the Fahrenheit scale it is \(\frac{5}{9}\) times that in Celsius and Kelvin scales.
2. If is assumed during the derivations of (2) and (4) that the value of the coefficient of expansion of a liquid is the same for all ranges of temperature. Precise observations show that the value changes, though the changes are very small. Hence, the values deduced above are the average values of γ and γ’, for the temperature range between t1 and t2. However, in practice, the values of γ and γ’ of a liquid are taken as constants for all temperature ranges.
3. While defining γ or γ’, initial volume at any temperature is taken. But for finer measurements, volume at 0°C should be taken as the initial volume. In practice, the difference is ignored.
Values of the coefficient of real expansion of a few liquids are shown in the following table.
Coefficient of real expansion
Relation between the Coefficients of Apparent and Real Expansions of Liquids
Let the volume of certain mass of a liquid in a container at a temperature t1 = V1. At a temperature t2, the apparent volume of that liquid = V’2 and its real volume = V2.
The part of the container, that contained the liquid at t1, has a volume V at t2.
∴ Apparent expansion of the liquid = V2‘ – V1 and real expansion = V2 – V1.
Expansion of the part of the container containing the liquid at t1 = V – V1
Since for a liquid, real expansion = apparent expansion + expansion of container,
∴ V2 – V1 = (V2‘ – V1) + (V – V1).
Dividing both sides by V1t (where t = t2 – t1),
\(\frac{V_2-V_1}{V_1 t}\) = \(\frac{V_2{ }^{\prime}-V_1}{V_1 t}\) + \(\frac{V-V_1}{V_1 t}\)
or, γ = γ’ + γg [γg = coefficient of volume expansion of the material of the container]
Hence, the coefficient of real expansion of a liquid = the coefficient of apparent expansion of the liquid + the coefficient of volume expansion of the material of the container.
Relation between Density and Coefficient of Real Expansion of Liquids
The volume of a liquid increases with the increase in tem-perature. Thus the density decreases. Water between 0°C and 4°C is an exception, and that will be discussed later.
Let the mass of some liquid be m, the volume of that liquid be V1 and the density be ρ1, at temperature t1.
At temperature t2, its volume becomes V2 and density ρ2. Consider t2 > t1.
Hence, m1 = V1ρ1 = V2ρ2 or, \(\frac{\rho_1}{\rho_2}\) = \(\frac{V_2}{V_1}\) ……… (1)
If the coefficient of real expansion of the liquid is γ, then V2 = V1{1 + γ(t2 – t1)}
V2 = V1{1 + γ(t2 – t1)}
or, \(\frac{V_2}{V_1}\) = 1 + γ(t2 – t1) ……….. (2)
From equations (1) and (2) we get,
\(\frac{\rho_1}{\rho_2}\) = {1 + γ(t2 – t1)}
or, ρ1 = ρ2[1 + γ(t2 – t1)] ….. (3)
∴ ρ2 = \(\frac{\rho_1}{1+\gamma\left(t_2-t_1\right)}\) = ρ1{1 + γ(t2 – t1)}-1
= ρ1[1 – γ(t2 – t1)] …….. (4)
neglecting higher powers of γ, as it is very small.
Hence, the density of a liquid decreases with the increase in temperature.
Equations (3) and (4) both give the relation between the coefficient of real expansion and the density of the liquid.
Equation (4) can be written as γ = \(\frac{\rho_1-\rho_2}{\rho_1\left(t_2-t_1\right)}\).
Thus if densities of a liquid at two different temperatures are known, its coefficient of real expansion can be found out.
Numerical Examples
Example 1.
Coefficient of apparent expansion of mercury with respect to glass is 153 × 10-6°C-1. Find the coeffi-cient of linear expansion (αg) of glass where coefficient of real expansion of mercury is 180 × 10-6°C-1.
Solution:
Since coefficient of real expansion (γ) of mercury = coefficient of apparent expansion (γ’) of mercury + coefficient of volume expansion of the material of the container (γg),
180 × 10-6 = 153 × 10-6 + γg
or, γg = (180 – 153) × 10-6 = 27 × 10-6°C-1
Also, γg = 3αg [αg = coefficient of linear expansion of the material of the container]
∴ αg = \(\frac{\gamma_g}{3}\) = \(\frac{27}{3}\) × 10-6 = 9 × 10-6°C-1.
Example 2.
A liquid has a coefficient of apparent expansion, 18 × 10-5°C-1 for an iron container, and 14.46 × 10-5°C-1 for an aluminium container. If the coefficient of linear expansion of aluminium is 2.38 × 10-5°C-1, find that of iron.
Solution:
We know γ = γ’ + γg, where, γ = coefficient of real expansion of the liquid, γ’ = coefficient of apparent expansion of the liquid, γ’ = coefficient of volume expansion of the material of the container.
γ = 18 × 10-5 + γiron ……. (1)
and for aluminium container,
γ = 14.46 × 10-5 + γAl
= 14.46 × 10-5 + 3 × 2.38 × 10-5°C-1
[∵ γAl = 3αAl]
= 21.60 × 10-5°C-1
From (1) and (2), we get,
γiron = 21.60 × 10-5 – 18 × 10-5 = 3.6 × 10-5
or, 3αiron = 3.6 × 10-5°C-1
∴ αiron = 1.2 × 10-5C-1
Example 3.
A thin long glass tube of uniform cross-section contains 1 m long mercury thread at 0°C. At 100°C, the mercury thread increases by 16.5 mm. If the coefficient of real expansion of mercury is 0.000182 °C-1, find the coefficient of linear expansion of glass.
Solution:
Let the areas of cross-section of the glass tube be A0 cm2 and A100 cm2 at 0°C and 100°C respectively.
Volume of mercury at 0°C, V0 = 1oo × A0 cm3 ……. (1)
and volume of mercury at 100°C,
V100 = 101.65 × A10o cm3 …….. (2)
Also from surface expansion of glass we get,
A100 = A0(1 + 100β)
= A0(1 + 100 × 2αg) …… (3)
(where αg = coefficient of linear expansion of glass]
From definition, γ for mercury = \(\frac{V_{100}-V_0}{V_0 \times 100}\) …….. (4)
Substituting the values of γ, V0, V1 and using (3), we have,
Example 4.
The Internal volume of a glass flask is V cm3. What volume of mercury should be kept in the flask so that the volume of the empty space over mercury in the flask remains constant at all temperatures? Coefficient of volume expansion of mercury = 1.8 × 10-4°C-1 and coefficient of linear expansion of glass = 9 × 10-6°C-1.
Solution:
Let the required volume of mercury be x cm3.
To keep the volume of empty space constant at all temperatures, the expansion of mercury should be equal to that of glass for the same rise in temperature. If the temperature rise is t°C,
x × 1.8 × 10-4 × t = V × 3 × 9 × 10-6 × t
or, x = \(\frac{V \times 27 \times 10^{-6}}{1.8 \times 10^{-4}}\) = \(\frac{27}{180}\)V = \(\frac{3}{20}\)V
∴ \(\frac{3}{20}\)th part of the flask should be filled up with mercury.
Example 5.
The volume expansion coefficients of glass and mercury are 2.4 × 10-5°C-1 and 1.8 × 10-4°C-1 respectively. What volume of mercury should be kept in the flask so that the volume of the empty space over mercury in the flask remains constant at all temperatures?
Solution:
Let the required volume of mercury = x cm3 and
volume of the glass container = V cm3.
To keep the volume of empty space constant at any temperature, the expansion of mercury should be equal to that of the glass container for the same rise in temperature.
If the temperature rise is t°C,
x × 1.8 × 10-4 × t = V × 2.4 × 10-5 × t
or, x = \(\frac{V \times 2.4 \times 10^{-5}}{1.8 \times 10^{-4}}\) = V × \(\frac{24}{180}\) = \(\frac{2}{15}\)V
∴ \(\frac{2}{15}\)th part of the flask should be filled with mercury.
Example 6.
The Internal volume of a glass flask is 540 cm3. What volume of mercury should be kept in the flask so that the volume of the empty space remains constant at any temperature? Real expansion of mercury = 1.8 × 10-4°C-1 and volume expansion of glass = 2.5 × 10-5°C-1.
Solution:
Let the volume of mercury be x cm3.
To keep the volume of the empty space constant at any temperature. the expansion of mercury should he equal to that of the glass flask for the same rise in temperature. If the temperature rise is t°C, then
x × 1.8 × 10-4 × t = 540 × 2.5 × 10-5 × t or, x = 75
∴75 cm3 of mercury should be kept in the flask.
Example 7.
The volume of the bulb of a mercury thermometer is 1 cm3 at 0°C, and it is filled with mercury at that temperature. The tube attached to the bulb has an area of cross-section 0.1 mm2. If the coefficient of apparent expansion of mercury is 16 × 10-5 °C-1, find the length up to which mercury expands in the tube, when the bulb is immersed in boiling water.
Solution:
The expansion of mercury in the bulb = initial volume of mercury × its coefficient of apparent expansion × increase in temperature
= 1 × 16 × 10-5 × (100 – 0) = 16 × 10-3 cm3
Suppose the mercury thread expands by a length x in the tube.
∴ x × 0.1 × 10-2 = 16 × 10-3 ∴ x = 16 cm.
Example 8.
1 g water has a volume of 1 cm3 at 4°C. Its volume at 60°C is 1.0169 cm3. Calculate the average coefficient of real expansion of water between these two temperatures.
Solution:
Coefficient of real expansion (average value),
γ = \(\frac{V_2-V_1}{V_1\left(t_2-t_1\right)}\)
= \(\frac{1.0169-1}{1 \times(60-4)}\) = \(\frac{0.0169}{56}\) = 3.02 × 10-4°C-1.
Example 9.
Masses of 10 cm3 of water are 9.998 g and 10 g at 0°C and 4°C respectively. Find the average coefficient of real expansion of water within the range of these temperatures.
Solution:
Density of water at 0°C,
ρ0 = \(\frac{9.998}{10}\) = 0.9998g ᐧ cm-3.
Density of water at 4°C, ρ4 = \(\frac{10}{10}\) = 1 g ᐧ cm-3
Therefore, the average coefficient of real expansion of water within the range of 0°C to 4°C,
γ = \(\frac{\rho_0-\rho_4}{\rho_0 \times t}\) = \(\frac{0.9998-1}{0.9998 \times(4-0)}\) = -5 × 10-5°C-1
The negative value of γ indicates that water contracts when it is heated from 0°C to 4°C.
Example 10.
Density of mercury at 0°C is 13.5955 g ᐧ cm-3. If the coefficient of linear expansion of mercury is 0.000061°C-1, find its density at 60°C.
Solution:
Given, ρ0 = 13.5955 g ᐧ cm-3, ρ60 = ?
γ = 3α = 3 × 61 × 10-6°C-1
Density of mercury at 60°C,
ρ60 = \(\frac{\rho_0}{1+\gamma t}\) = \(\frac{13.5955}{1+3 \times 61 \times 10^{-6} \times 60}\) [∵ t = 60° – 0° = 60°]
= \(\frac{13.5955}{1+10980 \times 10^{-6}}\) = 13.448 g ᐧ cm-3.