Animated Solution for Physics - Thermodynamics: The volume $V$ of a given mass of monoatomic gas changes with temperature $T$ according to the relation $V=k{T}^{2/3}$. The work done when temperature changes by $90\text{ K}$ will be $xR$. The value of $x$ is……… . $[R=\text{universal gas constant}]$

The beauty of thermodynamics lies in its ability to connect macroscopic, observable quantities—like pressure, volume, and temperature—through elegant mathematical relationships. In this problem, we are presented with a monoatomic gas undergoing a specific thermodynamic process where its volume $V$ is directly proportional to the temperature $T$ raised to the power of $2/3$. Our mission is to calculate the work done by the gas when its temperature increases by $90\text{ K}$.

Let's embark on this mathematical journey and see how calculus helps us unravel the physical reality of the gas.

We are given the relation: where $k$ is a constant of proportionality. We are also given the change in temperature, $\Delta T=90\text{ K}$. The work done is expressed as $W=xR$, and we need to find the numerical value of $x$.

Since the problem mentions "a given mass" without specifying the exact amount, it is a standard convention in such competitive exam problems to assume the number of moles $n=1$ for the sake of determining the coefficient $x$.

In thermodynamics, the work done $W$ by a gas during expansion or compression is given by the integral of pressure with respect to volume:

However, we have a slight roadblock. Our given relation is between volume $V$ and temperature $T$, but our integral involves pressure $p$ and volume $V$. To proceed, we must express pressure in terms of variables we know. Enter the ideal gas law:

Rearranging this for pressure, we get:

Substituting this into our work integral gives us:

Our integral now contains three variables: $T$, $V$, and $dV$. To evaluate it, we must transform it into an integral with respect to a single variable. Since we are given the change in temperature $\Delta T$, it makes perfect sense to convert everything into terms of $T$.

We need to find an expression for $dV$ in terms of $dT$. We do this by differentiating our given volume-temperature relation with respect to $T$:

Using the power rule of differentiation, we get:

Multiplying both sides by $dT$, we isolate $dV$:

Now, we substitute our expressions for $V$ and $dV$ back into the work integral. This is where the magic happens:

At first glance, this looks like a messy algebraic soup. But let's carefully simplify it. Notice how the constant $k$ appears in both the numerator and the denominator. It perfectly cancels out! We can also pull the constants $\frac{2}{3}$, $n$, and $R$ outside the integral:

Let's combine the powers of $T$ in the numerator. $T^1\cdot T^{-1/3}=T^{1-1/3}=T^{2/3}$.

So, the integral becomes:

The $T^{2/3}$ terms cancel out completely! We are left with the simplest integral possible:

Evaluating this integral from an initial temperature $T_1$ to a final temperature $T_2$ yields:

Since $T_2-T_1$ is simply the change in temperature $\Delta T$, we arrive at a beautifully concise formula for the work done in this specific process:

We have successfully distilled the complex calculus into a simple algebraic formula. Now, we just plug in the given values. Assuming $n=1$ mole and using the given $\Delta T=90\text{ K}$:

The problem states that the work done is $xR$. By comparing our result, we can clearly see that:

$$x=60$$

This problem reveals a powerful generalized pattern. If a gas undergoes a process where $V=k{T}^m$, following the exact same calculus steps will show that the work done is always $W=m\cdot nR\Delta T$. In our case, $m=2/3$, which immediately gives $W=\frac{2}{3}nR\Delta T$. Recognizing these patterns can save you immense time in competitive exams!