Question

For 1 mol of gas, the plot of pV vs. p is shown below. p is the pressure and V is the volume of the gas. What is the value of compressibility factor at point A?

• A. \(1-\frac{a}{RTV}\)
• B. \(1+\frac{b}{V}\)
• C. \(1-\frac{b}{V}\)
• D. \(1+\frac{a}{RTV}\)

Hint:

When solving compressibility factor problems: 
Start by expressing \(Z = \frac{PV}{RT}\) and substitute the appropriate gas law (e.g., Van der Waals equation). 
Expand and simplify using the given conditions to isolate terms involving \(V\), \(a\), and \(b\). 
Pay attention to the behavior of \(Z\) in real gas scenarios.

Answer 1

The Correct Option is A

The problem involves determining the compressibility factor Z for a gas, which is a measure of deviation from ideal gas behavior. The compressibility factor is given by the formula:

Z = \frac{pV}{nRT}

For 1 mol of gas, n = 1, the equation simplifies to:

Z = \frac{pV}{RT}

The typical equation for real gases is the Van der Waals equation:

\left(p + \frac{a}{V^2}\right)(V - b) = RT

The compressibility factor can be expressed in terms of Van der Waals constants a and b as:

• Z = 1 - \frac{a}{RTV} + \frac{b}{V}

At point A in the given plot, we need to evaluate the compressibility factor. The option 1-\frac{a}{RTV} indicates the deviation due to attraction between molecules. This correctly reflects a situation where attractive forces are significant, thus the gas deviates negatively from ideal behavior.

Checking other options:

• 1+\frac{b}{V} suggests positive deviation primarily due to volume exclusion, which may not be the strongest factor at point A.
• 1-\frac{b}{V} and 1+\frac{a}{RTV} are not standard forms or typical for assessing compressibility factor based on the given context.

Hence, the correct answer is 1-\frac{a}{RTV} as it aligns with the behavior expected at point A, where attractive forces are prominently affecting the gas's deviation from ideal conditions.