Derivation of Apparent Expansion of Liquid The formula for apparent expansion of liquid in a container is given as follows: 
$$\triangle V=V_0{(\gamma_{l}-\gamma_{c}})\triangle T$$ Where $V_0$ is the volume of the liquid as well as the container in which the liquid is filled and $\gamma_l$ is the coefficient of cubical expansion of liquid and the $\gamma_c$ is the coefficient of cubical expansion of the container. 
The final volume of liquid is calculated as : $V_{l_{f}}=V_0*\gamma_l*\triangle T$ and the final volume of container is given by : $V_{c_{f}}=V_0*\gamma_c*\triangle T$. What confuses me  is the fact that $V_0$ is used in both the equations instead of $V_{c_{inital}}$ or $V_{l_{inital}}$ . Please explain...
 A: I'm assuming that you don't have any trouble with $V_{l_{f}}=V_0*\gamma_l*\triangle T$ but rather with $V_{c_{f}}=V_0*\gamma_c*\triangle T$.
If you are literally thinking of the volume of the container it will be different than the volume of liquid it holds e.g. the volume of glass that makes up a beaker will be different than the volume of water in the beaker.  But $V_c$ doesn't refer to the volume of glass that makes up the beaker it refers to the volume of glass that is missing from a solid hunk to glass so that there's room to put the water in.  The key is when the glass beaker expands it expands as if it were a solid hunk of glass.  The space left by the missing glass (where the water is) expands in the same way that a block of glass of that size and shape would expand.
Think of the classic thermal expansion demonstration with the brass ball and ring.  People often think when the ring is heated the inside diameter of the hole will shrink as the metal expands but in fact the hole expands in the same way that a piece of brass of the same size and shape of the hole would expand.
