Animated Solution for Physics - Thermodynamics: An ideal gas is expanding such that $p{T}^3=\text{constant}$. The coefficient of volume expansion of the gas is

A Unique Expansion

Imagine a gas expanding, but not in a standard way. It follows a unique thermodynamic process governed by the equation $p{T}^3=\text{constant}$. Our mission is to find its coefficient of volume expansion, denoted by $\gamma$.

Defining Gamma

Before we dive into the algebra, we must anchor ourselves to the fundamental definition. What exactly is the coefficient of volume expansion? Physically, it represents the fractional change in volume per unit change in temperature. Mathematically, it is expressed as:

To conquer this problem, our primary objective is to find the derivative $\frac{dV}{dT}$. This means we need to express the volume $V$ purely as a function of the temperature $T$.

Eliminating Pressure

Our given equation, $p{T}^3=\text{constant}$, contains pressure $p$, which is an unwanted guest in our quest to relate $V$ and $T$. To evict it, we call upon the trusty ideal gas law:

Substituting this expression for $p$ back into our process equation, we get:

Now, let's rearrange this to isolate $V$. Multiplying $T$ by $T^3$ gives us $T^4$. Moving $V$ to the other side yields:

Since $n$, $R$, and $C$ are all constants, their combination is also a constant. Let's wrap them up into a neat little package called $k$:

Finding the Rate of Change

With $V$ beautifully isolated as a function of $T$, we are ready to unleash calculus. Differentiating both sides with respect to $T$ using the power rule, we find:

This derivative represents the slope of the tangent on a $V-T$ graph. However, our final answer shouldn't depend on the arbitrary constant $k$. We must eliminate it by substituting $k=\frac{V}{T^4}$ back into our derivative:

Simplifying this, the $T^3$ cancels out most of the denominator, leaving us with a clean and elegant expression:

A Beautiful Cancellation

We are at the finish line! Let's bring back our master equation for $\gamma$ and substitute our newly found derivative:

Notice how the volume $V$ in the numerator and denominator perfectly cancel each other out. The physics is telling us that the fractional expansion rate depends only on the temperature! We are left with our final answer:

To put this into perspective, for a standard isobaric (constant pressure) process, volume is directly proportional to temperature ($V\propto T$), which gives a coefficient of $\frac{1}{T}$. But in our unique process, the volume grows with the fourth power of temperature ($V\propto T^4$). It expands four times as aggressively, which is exactly why our coefficient is $\frac{4}{T}$!