Animated Solution for Physics - Thermodynamics: An ideal gas is expanding such that $p{T}^2=\text{constant}$. The coefficient of volume expansion of the gas is

The behavior of ideal gases under various thermodynamic processes is one of the most fascinating areas of physics. In this problem, we are given a unique process equation and asked to find the coefficient of volume expansion. Let's embark on this mathematical journey and uncover the hidden relationship between volume and temperature!

Before we dive into the algebra, we must understand what we are looking for. The coefficient of volume expansion, denoted by the Greek letter $\gamma$, measures the fractional change in volume per degree change in temperature. Mathematically, it is defined as:

To evaluate this derivative, we need to express the volume $V$ purely as a function of the temperature $T$. However, our given process equation is $p{T}^2=\text{constant}$. It contains pressure $p$, which we need to eliminate.

Eliminating Pressure

How do we get rid of pressure? We call upon the trusty ideal gas equation! For any ideal gas, we know that:

By rearranging this, we can express pressure in terms of volume and temperature:

Now, let's substitute this expression for $p$ back into our original process equation:

Let's clean up this equation. Multiplying the terms in the numerator gives us:

Here is a crucial realization: the number of moles $n$ and the universal gas constant $R$ are themselves constants. If we divide the constant on the right side by $nR$, we just get another constant! Let's call this new constant $k$.

We can rewrite this elegantly as:

This tells us that the volume $V$ is directly proportional to the cube of the temperature $T$.

Logarithmic Differentiation

We need to find $\frac{dV}{dT}$. While we could isolate $V$ and use the power rule, there is a much more elegant technique: logarithmic differentiation. Let's take the natural logarithm ($\ln$) of both sides of our equation:

Using the properties of logarithms, we can expand the left side:

Now, let's differentiate this entire equation with respect to temperature $T$. The derivative of $\ln T$ is $\frac{1}{T}$, and by the chain rule, the derivative of $\ln V$ is $\frac{1}{V}\frac{dV}{dT}$. The derivative of a constant is simply zero.

Look closely at the equation we just derived. The term $\frac{1}{V}\frac{dV}{dT}$ is exactly the definition of our coefficient of volume expansion, $\gamma$! Let's isolate it by moving it to the right side of the equation:

Therefore, we have found our answer:

This beautiful result tells us that for this specific thermodynamic process, the coefficient of volume expansion is inversely proportional to the temperature. As the gas gets hotter, its fractional expansion per degree of heating actually decreases. Physics is truly elegant when viewed through the lens of mathematics!