- Note
- Things to remember
- Videos

**Construction**

Absolute expansivity of liquid is determined by Dulong and Petit’s experiment as shown in figure A. It consist of U-shaped ABCD tube having a uniform diameter. The liquid whose expansivity is to be measured inside the tube the limb Jacket (J_{1}) and limb CD is covered with Jacket (J_{2}). There is cold water circulating system in J_{1 }to cool liquid inside tube AB and steam circulating system in J_{2} to heat liquid inside the tube CD. The thermometer T_{1 }and T_{2} are placed in J_{1 }and J_{2 }to measure the temperature J_{1 }and J_{2}. The horizontal portion of tube BC is wrapped with wet cloth in order to prevent the flow of heat from one limb to another.

**Working**

In the experiment, the thermometer T_{1 }and T_{2 }will show the constant temperature when the steady temperature is reached. Suppose, \(\rho_1 \text{and} \rho_2\)be the densities of liquid h_{1} and h_{2} be the height of liquid in tube AB and CD in equilibrium position

Since,

B and C are in same horizontal level at equilibrium

Pressure at B = Pressure at C

$$\text{or, } h_1\rho_1 g = h_1 \rho_2 g \dots (i)$$

Since,

$$\rho_1 = \rho_2 [1 + \gamma (\theta_2 - \theta_1)]\dots (ii)$$

Using (ii) in (i)

$$\text{or, } h_1\rho_2 [1 + \gamma (\theta_2 - \theta_1)] = h_2 \rho_2 $$

$$\text{or, } h_1 + h_1 \gamma (\theta_2 - \theta_1)] = h_2 $$

$$ \therefore \gamma = \frac{h_2 - h_1}{h_1(\theta_2 - \theta_1)}$$

From values of h_{1}, h_{2}, \(\theta_1 \text{and} \theta_2\) an absolute expansivity of the liquid is determined.

**Sources of errors**

- The wide separation of A and D makes the measurement of h
_{1}and h_{2} - The horizontal portion of the tube should be made a narrow otherwise transference of heat from hot to cold takes place which will affect the result.
- Surface tensions are different at a different temperature. So, it causes the difference of pressures across each free surface of mercury.
- The height of the mercury columns should be measured from the middle or axis of the horizontal portion of the tube.

In Fortin’s barometer, the brass scale is calibrated at 0^{o}Cand the mercury is adjusted at the same temperature. If the room temperature is greater than 0^{o}C, then both mercury and brass scale expand. As expansion in liquid and solid is different, correction in brass scale and in the density of mercury requires for the correct pressure at any temperature.

**Anomalous Expansion of Water**

When water at 0 ^{o}C is heated to 4^{o}C instead of increase in volume in contracts till 4^{o}C where its density is maximum at 4^{o}C and minimum volume at 4^{o}C but on further heating it expands normally. Water when cooled from 4^{o}C to 0^{o}C instead of contracting it expands. This peculiar behaviour of water is known as an anomalous expansion of water.

Hope verified experimentally the anomalous expansion of water.

**Construction**

The apparatus consists of a cylindrical jar containing water at about 10and surrounded by a jacket containing ice in the middle of the jar as shown in the figure. T_{1 }and T_{2 }are two thermometers to record the temperature of the water above and below the jacket respectively. Temperatures are recorded at an equal interval of time (say a half minute). A graph between the temperature along the y-axis and the time along the x-axis is drawn.

Initially, both thermometers record same temperature(room temperature). As time passes, T_{1 }still shows room temperature (10) but T_{2} shows 4^{o}C. After few minutes, T_{1} shows 0but T_{2} still shows 4^{o}C.

**Explanation**

When the water in the middle of the jar is cooled by the ice, its density increases and the lower thermometer starts showing the fall in the temperature till it reaches 4^{o}C as its density becomes maximum. When the water in the middle cools below 4^{o}C, its density decreases and these layers rise up. These results in the decrease in the temperature as shown by the thermometer T_{1}. It may show 0when ice is formed at the top. So, T_{1 }shows 0^{o}C and T_{2} shows 4^{o}C.

As the temperature increases to from 0, the reading given by brass scale is less than the correct reading given by mercury as shown in the figure. If h_{scale} is the reading given by brass scale of mercury and h_{true} height of mercury at \(\theta^{\circ}C\), then

$$h_\text{true} = h_\text{scale}(1 + \alpha\Delta\theta)\dots (i)$$

Where is linear expansivity of the brass scale, If be the density at. Then the atmospheric pressure at is

$$P = \rho_\theta h_\text{true} g \dots (ii)$$

The height of mercury column at 0^{o}Cwhich would exert equal pressure is called the corrected height h _{cor.}

$$P = \rho_0 h_\text{cor} g \dots (iii)$$

From equations (i) and (ii)

$$\rho_\theta h_\text{true} g=\rho_0 h_\text{cor} g $$

$$\rho_\theta h_\text{true} =\rho_0 h_\text{cor}$$

Let be the coefficient of the volume expansion of mercury then density of mercury

$$ \rho_\theta = \frac{\rho_0}{1 + \gamma\Delta\theta}$$

$$ \frac{\rho_0}{1 + \gamma\Delta\theta} h_\text{true} = \rho_0 h_\text{cor}$$

$$\text{or, }h_\text{true} = h_\text{cor}(1 + \gamma\Delta\theta)$$

Again from equation(i)

$$ h_\text{scale}(1 + \alpha\Delta\theta)= h_\text{cor}(1 + \gamma\Delta\theta)$$

$$ h_\text{cor} = h_\text{scale} \frac{(1 + \alpha\Delta\theta)}{(1 + \gamma\Delta\theta)}= h_\text{scale} (1 + \alpha\Delta\theta)(1 + \gamma\Delta\theta)^-1$$

Expanding by binomial theorem and neglecting higher powers of, we get

$$ h_\text{scale} = h_\text{cor}(1 + \alpha\Delta\theta)(1 - \gamma\Delta\theta) $$

$$= h_\text{cor} (1 + \alpha\Delta\theta - \gamma\Delta\theta - \gamma\alpha\Delta\theta^2) $$

Neglecting the products of and we have

$$ h_\text{cor} = h_\text{scale} (1 + \alpha\Delta\theta - \gamma\Delta\theta) $$

$$ = h_\text{scale} (1 + (\alpha - \gamma)\Delta\theta) $$

Hence, the correction depends on the difference between the cubical expansivity of mercury and expansivity of the scale.

In Fortin’s barometer, the brass scale is calibrated at 0^{o}Cand the mercury is adjusted at the same temperature.

When water at 0 ^{o}C is heated to 4^{o}C instead of increase in volume in contracts till 4^{o}C where its density is maximum at 4^{o}C and minimum volume at 4^{o}C but on further heating it expands normally.

The correction depends on the difference between the cubical expansivity of mercury and expansivity of the scale.

.-
## You scored /0

No discussion on this note yet. Be first to comment on this note