Animated Solution for Physics - Thermodynamics: In a certain thermodynamical process, the pressure of a gas depends on its volume as $k{V}^3$. The work done when the temperature changes from $100^{\circ }\text{C}$ to $300^{\circ }\text{C}$ will be ......... $nR$, where $n$ denotes number of moles of a gas.

Imagine you are observing a gas trapped inside a cylinder with a movable piston. As the gas is heated, it expands, but it doesn't just expand randomly. It follows a very specific, mathematically beautiful rule: its pressure $p$ is directly proportional to the cube of its volume $V$. We can write this as:

We are tasked with finding the work done by this gas as its temperature increases from $100^{\circ }\text{C}$ to $300^{\circ }\text{C}$. At first glance, this might seem tricky because work done is the integral of pressure with respect to volume ($W=\int pdV$), but we are given the limits in terms of temperature!

To bridge the gap between volume and temperature, we must call upon the fundamental law that governs all ideal gases—the Ideal Gas Equation:

Now, let's substitute our given process equation $p=k{V}^3$ into this ideal gas law. This will allow us to eliminate pressure and see how volume directly relates to temperature:

This is a powerful relation. It tells us exactly how the volume scales as the temperature changes. But we still need to find $pdV$.

Instead of solving for $V$ and dealing with messy fractional powers, let's use a neat calculus trick. We will differentiate both sides of our new equation $k{V}^4=nRT$ with respect to temperature $T$.

Applying the power rule to the left side, the derivative of $V^4$ is $4V^{3}dV$. Since $k$, $n$, and $R$ are constants, we get:

Look closely at the term in the parentheses. $k{V}^3$ is exactly our original pressure $p$! Let's substitute $p$ back into the equation:

Rearranging this gives us a direct, elegant expression for $pdV$ entirely in terms of $dT$:

We have successfully converted a complex volume integral into a trivial temperature integral. The work done $W$ is simply the integral of $pdV$:

Since $\frac{nR}{4}$ is a constant, it pulls out of the integral, leaving us with:

All that's left is to plug in the numbers. The change in temperature $\Delta T$ is the final temperature minus the initial temperature.

Remember, a change of $200^{\circ }\text{C}$ is exactly equivalent to a change of $200\text{ K}$. Substituting this into our work equation:

Thus, the coefficient we were looking for is 50.