# Real expansion , Apparent expansion of Liquid

Apparent expansion of Liquid in a Container

As liquids have no definite shape, the liner and superficial coefficient cannot be studied. Further, for heating it has to be kept in a container. Obviously when the liquid is heated, the container will also expand.

The observed expansion of liquid is called apparent expansion and it is different from real expansion of the liquid.

Initially container was completely filled with liquid. Temperature is increased by ΔT

Final volume of liquid VL = Vo(1 + γL ΔT )

Final volume of container VC = Vo(1 + γC ΔT )

So, overflow volume of liquid relative to container ΔV = VL – VC

ΔV = Vo( γL – γC)ΔT

So coefficient of apparent expansion of liquid w.r.t. container

γapparent = γL – γC

##### EXPANSION IN ENCLOSED VOLUME:

Increase in height of liquid level in tube when bulb was initially completely filled :

$\displaystyle h = \frac{apparent \; change \;in \; volume \; of \;liquid }{area \; of \; tube}$

$\displaystyle h = \frac{V_0(\gamma_L – \gamma_g)\Delta T}{A_0(1+2\alpha_g \Delta T)}$

γapparent = γL – γC

A sphere of metal A(γ1) just inside a spherical shell of metal B(γ2) at any temperature T. If the temperature of metals is increased and γ1 > γ2 then volume stress will be developed (compressive in metal A and tensile in metal B).

But if(γ1 < γ2) the free space will be created between metals.

With same amount of change in temp. the expansion of radius will be same

R = Ro(1 + αΔT )

Example : A one litre glass flask contains some mercury. It is found that at different temperature the volume of air inside the flask remains the same. What is the volume of mercury in this flask if coefficient of linear expansion of glass is while volume expansion of mercury is 1.8 x 10-4 /°C ?

Solution: If V is the volume of flask, VL of mercury and VA of air in it,

V = VL + VA

Now as with change in temperature volume of air remains constant, the expansion of mercury will be equal to that of the whole flask i.e.,

ΔV = ΔVL

V γG Δθ = VLγL Δθ

[ As , ΔV = Vγ Δθ ]

Here V = 1 litre = 1000 cc and γG = 3 α G = 27 × 10-6 / °C

VL = (1000 x 27 ×10-6) /(1.8 × 10-4 ) = 150 cc.

### Volume Coefficient :

Consider a given mass of a gas is heated through 1° C at constant pressure. Now the ratio of the increase in volume to the original volume at 0° C is defined as volume coefficient.

If VT and V0 be the volumes of a given mass of gas T° C and 0° C respectively, then

$\displaystyle \alpha_v = \frac{V_T -V_0}{V_0 T}$

Where αv is known as volume coefficient

### Pressure coefficient :

Consider a given mass of a gas is heated through 1° C at constant volume Now the ratio of the increase in pressure to the original pressure at 0° C is defined as the pressure coefficient. If PT and P0be the pressures of a given mass of a gas at T° C and 0°C respectively, then

$\displaystyle \alpha_p = \frac{P_T – P_0}{P_0 T}$

where αp is known as pressure coefficient.

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