Question

use the van der waals equation to answer the questions in the table below.
what are the units of ( a )?
what are the units of ( b )?
for carbon dioxide the numerical value of ( a ) is 3.592 and the numerical value of ( b ) is 0.0429.
use the van der waals equation to calculate the pressure of a sample of carbon dioxide at ( 60.0^{circ} mathrm{c} ) with a molar volume of ( 0.515 mathrm{~l} / mathrm{mol} ). round your answer to 3 significant digits.
use the ideal gas law to calculate the pressure of the same sample under the same conditions. round this answer to 3 significant digits also.

Explanation:

Step1: Determine units of \(a\)

The van der Waals equation is \((P + \frac{n^{2}a}{V^{2}})(V - nb)=nRT\). In the term \(\frac{n^{2}a}{V^{2}}\), the units of pressure \(P\) (in atm) must be the same as the units of \(\frac{n^{2}a}{V^{2}}\). Since \(n\) is in mol and \(V\) is in \(L\), solving for \(a\) gives \(a=\frac{PV^{2}}{n^{2}}\). So the units of \(a\) are \(\frac{L^{2}\cdot atm}{mol^{2}}\).

Step2: Determine units of \(b\)

In the term \(nb\) (from \(V - nb\)), \(V\) is in \(L\) and \(n\) is in mol. Solving for \(b\) gives \(b = \frac{V}{n}\), so the units of \(b\) are \(\frac{L}{mol}\).

Step3: Calculate pressure using van der Waals equation

First, convert temperature \(T = 60.0^{\circ}C=(60.0 + 273.15)K=333.15K\).
The van der Waals equation for 1 - mole (\(n = 1\)) is \(P=\frac{RT}{V - b}-\frac{a}{V^{2}}\).
Substitute \(R = 0.0821\frac{L\cdot atm}{mol\cdot K}\), \(a = 3.592\frac{L^{2}\cdot atm}{mol^{2}}\), \(b=0.0429\frac{L}{mol}\), \(V = 0.515\frac{L}{mol}\), \(T = 333.15K\) into the equation:
\(P=\frac{0.0821\times333.15}{0.515 - 0.0429}-\frac{3.592}{(0.515)^{2}}\)
\(P=\frac{27.351615}{0.4721}-\frac{3.592}{0.265225}\)
\(P = 57.936 - 13.543\)
\(P\approx44.4atm\)

Step4: Calculate pressure using Ideal Gas Law

The Ideal Gas Law is \(PV=nRT\) (for \(n = 1\)), so \(P=\frac{RT}{V}\).
Substitute \(R = 0.0821\frac{L\cdot atm}{mol\cdot K}\), \(T = 333.15K\), \(V = 0.515\frac{L}{mol}\)
\(P=\frac{0.0821\times333.15}{0.515}=\frac{27.351615}{0.515}\approx53.1atm\)

Answer:

Units of \(a\): \(\frac{L^{2}\cdot atm}{mol^{2}}\)
Units of \(b\): \(\frac{L}{mol}\)
Pressure (van der Waals): \(44.4atm\)
Pressure (Ideal Gas Law): \(53.1atm\)