Liquid Expansion in Expandable Containers

1. Real vs. Apparent Expansion

When you heat a liquid in a beaker, it's not just the liquid that expands. The beaker (the container) also expands. The change in the liquid level you observe is the net result of these two expansions.

This leads to two important definitions:

• Real Coefficient of Volume Expansion (${\gamma }_r$): This is the intrinsic property of the liquid itself, describing how much its volume changes per degree of temperature change, independent of any container.
  $\Delta V_{real}=V_0{\gamma }_r\Delta T$
• Apparent Coefficient of Volume Expansion (${\gamma }_a$): This describes the change in volume that we observe relative to the container. It's the real expansion of the liquid minus the expansion of the container.

To see the formal relationship between these coefficients, let's look at a short but clear explanation.

As the resource shows, the core relationship is:
$\Delta V_{real}=\Delta V_{apparent}+\Delta V_{vessel}$
Dividing by $V_0\Delta T$, we get the relationship between the coefficients:
${\gamma }_{real}={\gamma }_{apparent}+{\gamma }_{vessel}$
For problem-solving, it's often more intuitive to think about the apparent expansion coefficient as the difference between the liquid's and the container's coefficients:
${\gamma }_a={\gamma }_r-{\gamma }_v$
Remember that for a solid container made of an isotropic material with linear expansion coefficient ${\alpha }_v$, its volume expansion coefficient is ${\gamma }_v=3{\alpha }_v$.

2. Calculating the Volume of Overflow

A classic problem involves a container filled to the brim with a liquid. When heated, the liquid usually expands more than the solid container (${\gamma }_r>{\gamma }_v$), causing some of it to spill.

The volume that overflows is simply the difference between the liquid's real volume increase and the container's volume increase.

$\Delta V_{overflow}=\Delta V_{real}-\Delta V_{vessel}$
$\Delta V_{overflow}=V_0{\gamma }_r\Delta T-V_0{\gamma }_v\Delta T$
$\Delta V_{overflow}=V_0({\gamma }_r-{\gamma }_v)\Delta T=V_0{\gamma }_a\Delta T$

The following video provides a crisp and clear derivation of this concept.

3. Key Problem Type: Constant Unoccupied Volume

Now for a more complex scenario frequently tested in JEE Advanced. Imagine a container that is only partially filled with a liquid. The question states that upon heating, the volume of the empty space above the liquid remains constant.

Let's break down the logic.

• Initial volume of the container: $V_C$
• Initial volume of the liquid: $V_L$
• Initial empty volume: $V_{empty}=V_C-V_L$

When heated by $\Delta T$, the new volumes are:

• New container volume: $V_C^{\prime }=V_C(1+{\gamma }_C\Delta T)$
• New liquid volume: $V_L^{\prime }=V_L(1+{\gamma }_L\Delta T)$
• New empty volume: $V_{empty}^{\prime }=V_C^{\prime }-V_L^{\prime }$

If the empty volume remains constant, then $V_{empty}^{\prime }=V_{empty}$.
$V_C^{\prime }-V_L^{\prime }=V_C-V_L$
$(V_C^{\prime }-V_C)=(V_L^{\prime }-V_L)$
$\Delta V_C=\Delta V_L$

This is the crucial condition: for the unoccupied volume to remain constant, the absolute increase in the container's volume must be equal to the absolute increase in the liquid's volume.

Substituting the expansion formulas:
$V_C{\gamma }_C\Delta T=V_L{\gamma }_L\Delta T$
This simplifies to a beautifully simple relationship:
$V_C{\gamma }_C=V_L{\gamma }_L$

This result is a powerful tool for solving this type of problem. Let's watch a video that explains this concept and then applies it directly to a past JEE problem.

The logic demonstrated in the video—equating the volume changes—is the key takeaway for this entire class of problems. Your ability to quickly identify this condition and apply the $V_C{\gamma }_C=V_L{\gamma }_L$ relationship will be a significant advantage.

Conclusion

Today, we extended our understanding of differential expansion from solids to liquid-container systems. The core idea remains the same: the observed effect is a result of two different components expanding at their own rates.

Key Takeaways:

• The expansion of a liquid in a container has two components: the real expansion of the liquid and the expansion of the vessel. The observed change is the apparent expansion.
• The coefficients are related by ${\gamma }_{real}={\gamma }_{apparent}+{\gamma }_{vessel}$.
• For a full container, the overflow volume is calculated as $\Delta V_{overflow}=V_0({\gamma }_{liquid}-{\gamma }_{vessel})\Delta T$.
• For a partially filled container, if the unoccupied volume is to remain constant upon heating, the change in the container's volume must equal the change in the liquid's volume: $\Delta V_C=\Delta V_L$, which simplifies to $V_C{\gamma }_C=V_L{\gamma }_L$.

Preview of the Next Lesson:
We will continue with the theme of thermal expansion, but focus on a very practical and common application: its effect on the accuracy of pendulum clocks. We'll see how a change in temperature affects the length of a pendulum, which in turn alters its time period, causing the clock to run fast or slow. This will be an interesting problem combining thermal physics with simple harmonic motion.