For one mole of a van der Waals gas when b = 0 and T=300K, the PV vs. 1/V plot is shown below. The v

Q: Why did we set $n=1$ in the van der Waals equation?

The question explicitly states 'For one mole of a van der Waals gas', which directly translates to $n=1$.

Q: What does $b=0$ physically mean in this context?

Setting $b=0$ means we are assuming the volume occupied by the gas molecules themselves is negligible compared to the total volume of the container. We are only focusing on the effect of intermolecular attractive forces ($a$).

Q: Why is the slope of the graph negative?

The negative slope occurs because the attractive forces, represented by '$a$', reduce the pressure exerted by the gas compared to an ideal gas. As density ($\frac{1}{V}$) increases, this attractive effect becomes stronger, lowering the overall $PV$ product.

Q: Can we find the temperature from the y-intercept of this graph?

Yes! The y-intercept of our rearranged equation is $RT$. If we extrapolate the graph to $\frac{1}{V} = 0$, we get $PV = RT$. We could use this intercept value to verify the given temperature $T=300K$.

Q: How do we determine the units of the constant '$a$'?

Look at the term $\frac{a}{V^2}$ which is added to Pressure ($P$). For the addition to be valid, $\frac{a}{V^2}$ must have units of pressure (atm). Since $V$ is in liters/mol, $a$ must have units of $\text{atm} \cdot (\text{liter/mol})^2$, which is $\text{atm liter}^2 \text{ mol}^{-2}$.

Question

For one mole of a van der Waals gas when b = 0 and T=300K, the PV vs. 1/V plot is shown below. The value of the van der Waals constant a (atm. liter² mol⁻²) is

Options:

(A) 1.0

(B) 4.5

(C) 1.5

(D) 3.0

Sigma Percentile · Accepted Answer

The behavior of real gases often deviates from the ideal gas law, and the van der Waals equation is our mathematical bridge to understanding these deviations. This problem is a beautiful example of how we can extract physical constants from graphical data.

Let's dive into the mechanics of this transformation.

### Analyzing the Setup

We are given a graph plotting $PV$ on the y-axis against $\frac{1}{V}$ on the x-axis.

The graph is a straight line with a negative slope. In physics and physical chemistry, whenever you see a straight line, your immediate instinct should be to map it to the equation of a straight line: $y = mx + c$.

Our goal is to manipulate the given gas equation to match this linear form.

### The Master Equation

The problem specifies one mole of a van der Waals gas. The standard equation is:

$\left(P + \frac{a}{V^2}ight)(V - b) = RT$

Here, **$a$** represents the magnitude of intermolecular attractive forces, and **$b$** represents the effective volume of the gas molecules.

The problem gives us a massive simplification: **$b = 0$**. This means we are assuming the molecules themselves occupy negligible volume. Substituting this into our equation gives:

$\left(P + \frac{a}{V^2}ight)V = RT$

### Algebraic Transformation

Now, we need to rearrange this equation so that $PV$ is isolated on one side, just like the 'y' in our straight-line equation. Let's expand the bracket:

$PV + \frac{a}{V} = RT$

Subtracting the second term from both sides, we get:

$PV = -a\left(\frac{1}{V}ight) + RT$

Look closely at this structure. It perfectly mirrors $y = mx + c$.

Our y-variable is $PV$, our x-variable is $\frac{1}{V}$, the y-intercept $c$ is $RT$, and crucially, the **slope $m$ is equal to $-a$**.

### Final Calculation

To find the van der Waals constant $a$, we simply need to calculate the slope of the line from the given graph.

We can pick two clear coordinates from the plot: $(2.0, 21.6)$ and $(3.0, 20.1)$. Using the standard slope formula $m = \frac{y_2 - y_1}{x_2 - x_1}$:

$m = \frac{20.1 - 21.6}{3.0 - 2.0}$

$m = \frac{-1.5}{1.0} = -1.5$

Since we established that the theoretical slope is $-a$, we can equate the two:

$-a = -1.5$

**$a = 1.5 	ext{ atm liter}^2 	ext{ mol}^{-2}$**

The negative slope physically indicates that as the gas becomes denser (higher $\frac{1}{V}$), the attractive forces ($a$) pull the molecules together, reducing the pressure they exert on the container walls compared to an ideal gas.

Animated Solution for Chemistry - States of Matter: For one mole of a van der Waals gas when b = 0 and T=300K, the PV vs. 1/V plot is shown below. The value of the van der Waals constant a (atm. liter² mol⁻²) is

Select Answer:

Visualized Solution

\text{Analyzing the Graph}

\text{van der Waals Equation}

\text{Applying Given Conditions}

\text{Rearranging the Equation}

\text{Comparing with Straight Line}

\text{Calculating Slope from Graph}

\text{Final Calculation}

\text{The Way Forward}

The Sigma Insight: Gaseous State

Analyzing the Setup

The Master Equation

Algebraic Transformation

Final Calculation

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