Animated Solution for Chemistry - States of Matter: A certain gas obeys $p(V_m-b)=RT$. The value of ${\left(\frac{\partial Z}{\partial p}\right)}_T$ is $\frac{xb}{RT}$. The value of $x$ is ............ . ($Z=$ compressibility factor)

Imagine a gas confined in a container. For an ideal gas, we use the familiar equation $pV=nRT$. However, real gases don't always behave perfectly. At very high pressures, the molecules are squeezed so closely together that their own physical volume becomes significant.

The equation given in our problem, $p(V_m-b)=RT$, is actually a high-pressure approximation of the famous van der Waals equation. Here, $V_m$ is the molar volume, and $b$ represents the "excluded volume"—the actual space taken up by the gas molecules themselves. Because the pressure is extremely high, the attractive forces between molecules (usually represented by the $\frac{a}{V_m^2}$ term) become negligible compared to the sheer force of the pressure, leaving us with this simplified equation.

To understand how much this real gas deviates from ideal behavior, we use the compressibility factor, denoted by $Z$. By definition, $Z=\frac{p{V}_m}{RT}$. For an ideal gas, $Z$ is exactly $1$.

Our goal is to find an expression for $Z$ using our given equation of state. Let's start by expanding the equation:

We want to isolate the $p{V}_m$ term, so we move $pb$ to the other side:

Now, to construct our compressibility factor $Z$, we divide the entire equation by $RT$:

Substituting $Z$ into the left side, we get a beautiful, clean expression:

This equation tells us a physical story: at high pressures, $Z$ is greater than $1$, meaning the gas is harder to compress than an ideal gas due to the repulsive forces and the finite size of the molecules ($b$).

The problem asks for the partial derivative of $Z$ with respect to pressure $p$, keeping temperature $T$ constant. This is written mathematically as ${\left(\frac{\partial Z}{\partial p}\right)}_T$.

Let's differentiate our expression for $Z$:

Since we are holding temperature $T$ constant, the entire term $\frac{b}{RT}$ acts as a constant multiplier. The derivative of the constant $1$ is $0$, and the derivative of $p$ with respect to $p$ is $1$.

The problem states that this derivative is equal to $\frac{xb}{RT}$. By comparing our derived result with the given expression, we can easily find $x$:

It is immediately clear that $x=1$. The elegance of this problem lies in manipulating a physical equation of state into a mathematical form that directly answers the question.