Question

You want to store \(165 \mathrm{g}\) of \(\mathrm{CO}_{2}\) gas in a \(12.5-\mathrm{L}\). tank at room temperature \(\left(25^{\circ} \mathrm{C}\right) .\) Calculate the pressure the gas would have using (a) the ideal gas law and (b) the van der Waals equation. (For \(\mathrm{CO}_{2}, a=3.59 \mathrm{atm} \cdot \mathrm{L}^{2} / \mathrm{mol}^{2}\) and \(b=\) \(0.0427 \mathrm{L} / \mathrm{mol} .)\)

Short answer

(a) Ideal pressure: 7.37 atm (b) Van der Waals pressure: 7.11 atm

Step-by-step solution

Convert Grams to Moles

First, we need to convert the mass of \(\mathrm{CO}_2\) from grams to moles using its molar mass. The molar mass of \(\mathrm{CO}_2\) is approximately 44.01 g/mol. \[\text{Number of moles} = \frac{165\, \mathrm{g}}{44.01\, \mathrm{g/mol}} \approx 3.75\, \mathrm{mol}.\]

Use the Ideal Gas Law

Next, we'll calculate the pressure using the Ideal Gas Law equation, \(PV = nRT\). We need to ensure that we use the correct units:- \(n = 3.75\, \mathrm{mol}\)- \(R = 0.0821\, \mathrm{L} \cdot \mathrm{atm} / (\mathrm{mol} \cdot \mathrm{K})\)- The temperature must be in Kelvin: \\[T = 25^\circ C + 273.15 = 298.15\, \mathrm{K}\]- Volume \(V = 12.5\, \mathrm{L}\) Now, solve for pressure P:\[P = \frac{nRT}{V} = \frac{(3.75 \times 0.0821 \times 298.15)}{12.5} \approx 7.37\, \mathrm{atm}\]

Understand the Van der Waals Equation

The Van der Waals equation accounts for non-ideal behavior of gases. It is expressed as:\[\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT\]Here, \(a = 3.59\, \mathrm{atm} \cdot \mathrm{L}^2 / \mathrm{mol}^2\) and \(b = 0.0427\, \mathrm{L} / \mathrm{mol}\). This equation will help us calculate more accurate pressure values.

Calculate Pressure Using Van der Waals Equation

Substitute the known values into the Van der Waals equation:\[\left(P + \frac{3.59 \cdot (3.75)^2}{12.5^2}\right)(12.5 - 3.75 \cdot 0.0427) = 3.75 \cdot 0.0821 \cdot 298.15\]First, calculate the correction term for pressure:\[\frac{an^2}{V^2} = \frac{3.59 \cdot (3.75)^2}{(12.5)^2} \approx 0.3225\, \mathrm{atm}\]Now calculate the corrected volume:\[V - nb = 12.5 - 3.75 \times 0.0427 \approx 12.34 \, \mathrm{L}\]Plug these into the equation:\[(P + 0.3225)(12.34) = 3.75 \times 0.0821 \times 298.15 \]Calculate the right side:\[3.75 \times 0.0821 \times 298.15 \approx 91.75\]Finally, solve for \(P\):\[P \cdot 12.34 + 0.3225 \cdot 12.34 = 91.75\]Solve for \(P\):\(P \cdot 12.34 \approx 91.75 - 3.98\)\[P \approx \frac{87.77}{12.34} \approx 7.11 \, \mathrm{atm}\]

Key concepts

van der Waals equation

The van der Waals equation is a modified version of the Ideal Gas Law and it accounts for the real-world behaviors of gases that the Ideal Gas Law overlooks. This equation is particularly useful when dealing with high pressures or low temperatures. It is expressed as \[\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT\].
Here, the terms \(a\) and \(b\) are constants specific to each gas. They correct the Ideal Gas Law for:

• Intermolecular attractions (adjusted by \(a\)), which reduce the pressure of the gas particles as they are pulled toward each other.
• The volumes of gas particles (adjusted by \(b\)), which indicate that gas particles take up space and thus the actual volume available to the particles is less than the total volume.

Given constants \(a = 3.59 \, \mathrm{atm} \, \cdot \, \mathrm{L}^2 / \mathrm{mol}^2\) and \(b = 0.0427 \, \mathrm{L} / \mathrm{mol}\) for \(\mathrm{CO}_2\), we can correct the assumed ideal gas behavior by modifying both pressure and volume values used to calculate the pressure experience due to these real-world interactions.

pressure calculation

Pressure calculation is an essential process in understanding how gases behave under different conditions. In the context of the Ideal Gas Law, pressure \(P\) is calculated using the formula \[P = \frac{nRT}{V}\]. In this formula:

• \(n\) represents the number of moles of gas,
• \(R\) is the ideal gas constant (often \(0.0821 \, \mathrm{L} \, \cdot \, \mathrm{atm} / (\mathrm{mol} \, \cdot \, \mathrm{K})\)),
• \(T\) is the temperature in Kelvin,
• \(V\) is the volume of the gas in liters.

For our exercise, use the calculated number of moles, the tank's volume, and the converted temperature in Kelvin to calculate pressure.
When using the van der Waals equation like for our given \(\mathrm{CO}_2\), the pressure is slightly different due to the included interactions and volume irregularities. It involves a bit more computation, including solving additional correction factors for both pressure and volume.

mole conversion

Converting grams to moles is a fundamental chemical process, essential for applying gas laws to real-world problems. Moles provide a way to count particles in chemistry, and knowing the molar mass of a substance allows us to convert between grams and moles. For \(\mathrm{CO}_2\), the molar mass is roughly \(44.01 \, \mathrm{g/mol}\).
To find the number of moles from a given mass, use the formula:\[\text{Number of moles} = \frac{\text{Mass in grams}}{\text{Molar mass in g/mol}}\]For example, converting \(165 \, \mathrm{g}\) of \(\mathrm{CO}_2\) involves dividing by the molar mass, giving approximately \(3.75 \, \mathrm{mol}\). This conversion lays the groundwork for using gas laws, like the Ideal Gas Law, to predict behavior like pressure under specified conditions.

gas laws

Gas laws are an array of equations that describe the behavior of gases, crucial for predicting how gases react under different conditions. Some fundamental gas laws include:

• Ideal Gas Law: Combines Charles's Law, Boyle's Law, and Avogadro's Law into one equation, \(PV = nRT\), useful for predicting the behavior of a gas assuming ideal conditions.
• van der Waals Equation: Adjusts the Ideal Gas Law by considering real gas behaviors and is especially useful at high pressures and low temperatures.

By using these laws, we can calculate important properties, like pressure and volume, and understand how temperature, volume, and molecular interactions influence gas behavior. In the real world, choosing the right gas law depends on the conditions under which the gas is studied, with the van der Waals equation suited for more complex, real scenario applications.