__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 V_{L} = V_{o}(1 + γ_{L }ΔT )

Final volume of container V_{C} = V_{o}(1 + γ_{C }ΔT )

So, overflow volume of liquid relative to container ΔV = V_{L} – V_{C}

ΔV = V_{o}( γ_{L} – γ_{C})ΔT

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

γ_{apparent} = γ_{L} – γ_{C}

__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)} $

γ_{apparent} = γ_{L} – γ_{C}

A sphere of metal A(γ_{1}) just inside a spherical shell of metal B(γ_{2}) at any temperature T. If the temperature of metals is increased and γ_{1} > γ_{2} then volume stress will be developed (compressive in metal A and tensile in metal B).

But if(γ_{1} < γ_{2}) the free space will be created between metals.

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

R = R_{o}(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, V_{L} of mercury and V_{A} of air in it,

V = V_{L} + V_{A}

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 = ΔV_{L}

V γ_{G} Δθ = V_{L}γ_{L} Δθ

[ As , ΔV = Vγ Δθ ]

Here V = 1 litre = 1000 cc and γ_{G} = 3 α _{G} = 27 × 10^{-6} / °C

V_{L} = (1000 x 27 ×10^{-6}) /(1.8 × 10^{-4} ) = 150 cc.

### Volume Coefficient & Pressure Coefficient of gases

__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 V_{T} and V_{0} 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 α_{v} 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 P_{T} and P_{0}be 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 α_{p} is known as pressure coefficient.