Question

A slight deviation from ideal behavior exists even at normal conditions. If it behaved ideally, 1 mol of \(\mathrm{CO}\) would occupy \(22.414 \mathrm{~L}\) and exert 1 atm pressure at \(273.15 \mathrm{~K}\). Calculate \(P_{\text {VDW }}\) for \(1.000 \mathrm{~mol}\) of \(\mathrm{CO}\) at \(273.15 \mathrm{~K} .\left(\right.\) Use \(\left.R=0.08206 \frac{\mathrm{atm} \cdot \mathrm{L}}{\mathrm{mol} \cdot \mathrm{K}}\right)\)

Short answer

The corrected Van der Waals pressure is 0.9988 atm.

Step-by-step solution

Write down the given values

Given values are: \( n = 1 \text{ mol} \), \( V = 22.414 \text{ L} \), \( T = 273.15 \text{ K} \), and \( R = 0.08206 \frac{\text{atm} \times \text{L}}{\text{mol} \times \text{K}} \).

Recall the Van der Waals Equation

The Van der Waals equation for real gases is: \[ \biggl( P + \frac{n^2a}{V^2} \biggr) (V - nb) = nRT \] where: \( P \) = pressure, \( a \) = Van der Waals constant for intermolecular forces, \( b \) = Van der Waals constant for volume.

Identify Van der Waals constants for CO

For \( \text{CO} \), the Van der Waals constants are: \( a = 1.49 \frac{\text{L}^2 \text{atm}}{\text{mol}^2} \) and \( b = 0.0391 \frac{\text{L}}{\text{mol}} \).

Substitute values into the Van der Waals equation

Substitute the values into the equation: \[ \biggl( P + \frac{(1 \text{ mol})^2 \times 1.49 \frac{\text{L}^2 \times \text{atm}}{\text{mol}^2}}{(22.414 \text{ L})^2} \biggr) (22.414 \text{ L} - 1 \text{ mol} \times 0.0391 \frac{\text{L}}{\text{mol}}) = (1 \text{ mol}) \times 0.08206 \frac{\text{atm} \times \text{L}}{\text{mol} \times \text{K}} \times 273.15 \text{ K} \]

Simplify the equation

First, calculate the correction term for pressure: \[ \frac{1 \times 1.49}{22.414^2} = 0.00296 \text{ atm} \] Then the effective volume: \[ 22.414 \text{ L} - 0.0391 \text{ L} = 22.3749 \text{ L} \] Now, rewrite the equation: \[ \biggl( P + 0.00296 \biggr) \times 22.3749 = 1 \times 0.08206 \times 273.15 \]

Solve for P

Calculate the right-hand side: \[ 0.08206 \times 273.15 = 22.4148 \text{ atm} \times \text{L} \] Then solve for \( P \): \[ (P + 0.00296) \times 22.3749 = 22.4148 \] Therefore: \[ P + 0.00296 = \frac{22.4148}{22.3749} = 1.0018 \text{ atm} \] Finally: \[ P = 1.0018 - 0.00296 = 0.9988 \text{ atm} \]

Key concepts

real gases

In the study of gases, 'real gases' refer to actual gases that do not behave perfectly as predicted by the ideal gas law. This discrepancy arises from the presence of intermolecular forces and the actual volume occupied by gas molecules. Under many conditions, real gases approximate the behavior of an ideal gas, particularly at high temperatures and low pressures. However, at other conditions such as high pressures or low temperatures, deviations become significant, and more complex models like the Van der Waals equation, which accounts for these factors, become necessary. It's crucial to understand how real gases differ to accurately predict their behavior under various conditions.

non-ideal behavior

The 'non-ideal behavior' of gases features deviations from the ideal gas law (PV=nRT). This law assumes no intermolecular attractions and that gas particles occupy no volume. In reality, gas particles do attract each other and do occupy a finite volume, leading to deviations especially noticeable at high pressures and low temperatures. Non-ideal behavior is addressed by various models, most notably the Van der Waals equation, which modifies the ideal gas law by incorporating constants that adjust for these interactions. This helps to create a more accurate representation of the behavior of real gases, allowing for better predictions in practical applications.

ideal gas law

The 'ideal gas law' is an equation that relates the pressure (P), volume (V), temperature (T), and number of moles (n) of a gas, expressed as PV=nRT, where R is the universal gas constant. This law assumes that gas molecules do not interact and occupy no space. It works well under many conditions, particularly at low pressures and high temperatures. However, deviations occur mostly due to intermolecular forces and the finite size of gas particles, leading to the need for corrections which are provided by equations for real gases, like the Van der Waals equation. Despite its limitations, the ideal gas law remains a fundamental principle in understanding and calculating gas behavior.

intermolecular forces

Intermolecular forces are the forces of attraction and repulsion between molecules. These forces significantly affect the behavior of real gases. In the context of the Van der Waals equation, the constant 'a' represents the strength of these attractive forces. In an ideal gas, such forces are assumed negligible, but in reality, they cause gases to deviate from ideal behavior. These interactions become especially significant at high pressures and low temperatures where molecules are closer together. By incorporating these forces into calculations, we can more accurately predict the properties and behaviors of real gases.