Problem 88 - Chemistry Central Science

Question

Calculate the pressure that \(\mathrm{CCl}_{4}\) will exert at \(40^{\circ} \mathrm{C}\) if \(1.00 \mathrm{~mol}\) occupies \(28.0 \mathrm{~L}\), assuming that \((\mathrm{a}) \mathrm{CCl}_{4}\) obeys the ideal-gas equation; (b) \(\mathrm{CCl}_{4}\) obeys the van der Waals equation. (Values for the van der Waals constants are given in Table 10.3.) (c) Which would you expect to deviate more from ideal behavior under these conditions, \(\mathrm{Cl}_{2}\) or \(\mathrm{CCl}_{4} ?\) Explain.

Step-by-Step Solution

Answer

The calculated pressure of \(\mathrm{CCl}_4\) is approximately \(0.920\,\mathrm{atm}\) assuming ideal-gas behavior and \(0.875\,\mathrm{atm}\) assuming van der Waals behavior. Comparing the van der Waals constants of \(\mathrm{Cl}_2\) and \(\mathrm{CCl}_4\), we expect \(\mathrm{CCl}_4\) to deviate more from ideal behavior under the given conditions since it has larger values of a and b.

Step 1: Ideal Gas Law

First, we need to calculate the pressure of \(\mathrm{CCl}_{4}\) assuming it obeys the ideal-gas equation. The ideal-gas equation is: \[ PV = nRT \] Here, P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin. We have the volume (28.0 L) and number of moles (1.00 mol). We need to convert the given temperature from Celsius to Kelvin by adding 273.15 to the given temperature.

Step 2: Convert Temperature and Calculate Pressure

The temperature in Kelvin will be: \[ T = 40^{\circ}\mathrm{C} + 273.15 = 313.15 \,\mathrm{K} \] Now we have all the values needed to calculate the pressure of \(\mathrm{CCl}_{4}\) using the ideal-gas equation: \[ P = \frac{nRT}{V} = \frac{1.00\,\mathrm{mol} \times 0.0821\,\frac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}} \times 313.15\,\mathrm{K}}{28.0\,\mathrm{L}} \approx 0.920\,\mathrm{atm} \] So, the pressure exerted by \(\mathrm{CCl}_{4}\) assuming ideal-gas behavior is approximately \(0.920\,\mathrm{atm}\).

Step 3: Van der Waals Equation

Now, we calculate the pressure of \(\mathrm{CCl}_{4}\) assuming it obeys the van der Waals equation. The van der Waals equation is: \[ \left( P + a \frac{n^2}{V^2} \right) \left( V - nb \right) = nRT \] Where a and b are the van der Waals constants for the substance. From Table 10.3, the van der Waals constants for \(\mathrm{CCl}_{4}\) are \(a = 20.4\,\mathrm{L^2\cdot atm\cdot mol^{-2}}\) and \(b = 0.138\,\mathrm{L\cdot mol^{-1}}\). Now, we need to solve for P.

Step 4: Solve Van der Waals Equation for Pressure

Since we are given a, b, n, R, T, and V, we need to solve for P: \[ P = \frac{nRT}{V-nb} - a \frac{n^2}{V^2} \] And then calculate the pressure: \[ P = \frac{1.00\,\mathrm{mol} \times 0.0821\,\frac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}} \times 313.15\,\mathrm{K}}{(28.0\,\mathrm{L} - (1.00\,\mathrm{mol}\times0.138\,\mathrm{L\cdot mol^{-1}}))} - (20.4\,\mathrm{L^2\cdot atm\cdot mol^{-2}}) \frac{(1.00\,\mathrm{mol})^2}{(28.0\,\mathrm{L})^2} \approx 0.875\,\mathrm{atm} \] So, the pressure exerted by \(\mathrm{CCl}_{4}\) assuming van der Waals behavior is approximately \(0.875\,\mathrm{atm}\).

Step 5: Evaluate which molecule deviates more from ideal behavior

To decide which molecule, \(\mathrm{Cl}_{2}\) or \(\mathrm{CCl}_{4}\), is expected to deviate more from ideal behavior under the given conditions, we need to compare their van der Waals constants. From Table 10.3, we know the constants for both molecules: \(\mathrm{Cl}_{2}\): \(a = 6.49\,\mathrm{L^2\cdot atm\cdot mol^{-2}}\), \(b = 0.0562\,\mathrm{L\cdot mol^{-1}}\) \(\mathrm{CCl}_{4}\): \(a = 20.4\,\mathrm{L^2\cdot atm\cdot mol^{-2}}\), \(b = 0.138\,\mathrm{L\cdot mol^{-1}}\) Molecules with larger values of a and b are expected to deviate more from ideal behavior. Comparing the values, we can see that \(\mathrm{CCl}_{4}\) has larger values of both a and b compared to \(\mathrm{Cl}_{2}\). Hence, we expect \(\mathrm{CCl}_{4}\) to deviate more from ideal behavior under the given conditions.

Key Concepts

Van der Waals Equation

The Van der Waals equation is an essential tool for understanding real gases. It refines the Ideal Gas Law by correcting for the volume occupied by gas molecules and the forces between them. This is crucial for cases where gases do not behave ideally, especially under high pressure or low temperature.

The equation is \[ \left( P + a \frac{n^2}{V^2} \right) \left( V - nb \right) = nRT \] where:

• \(P\) is the pressure of the gas,
• \(V\) is the volume,
• \(n\) is the number of moles,
• \(R\) is the ideal gas constant (0.0821 L·atm/mol·K),
• \(T\) is the temperature in Kelvin,
• \(a\) and \(b\) are Van der Waals constants specific to each gas.

These constants include:

• \(a\) which corrects for intermolecular forces;
• \(b\) which accounts for the volume occupied by the gas molecules themselves.

When these constants are significant, the deviations from ideal gas behavior are notably pronounced, making the Van der Waals equation more applicable.

Pressure Calculation

Pressure calculation involves deriving the amount of force that the gas molecules exert on their container's walls. It is vital to accurately compute the pressure to predict the gas's behavior in various conditions.

For ideal gases, the pressure can be calculated using the formula: \[ P = \frac{nRT}{V} \]
However, when dealing with real gases, Van der Waals modifications are necessary. The adjusted pressure calculation becomes:\[ P = \frac{nRT}{V-nb} - a \frac{n^2}{V^2} \]
This formula reflects the influence of molecular size and interaction forces. Here's why this adjustment is critical:

• Molecular size: Real gas molecules occupy space, thereby reducing the effective volume available for movement.
• Molecular attraction: Attractions between molecules act to reduce the force with which they hit the walls, thus lowering the measured pressure.

By understanding these nuances, we can better anticipate how the gas will behave under different temperatures and pressures.

Deviations from Ideal Gas Behavior

Deviations from ideal gas behavior occur when the assumptions made in the ideal gas law are no longer valid. These assumptions include negligible molecular volume and no intermolecular forces. Real gases, particularly at high pressures and low temperatures, don’t always follow these assumptions.

To account for this, Van der Waals introduced corrections for:

• **Intermolecular forces:** In real gases, molecules experience attractions and repulsions. This changes how they interact with their container. For instance, attractive forces can lead to a lower-than-expected pressure.
• **Finite molecular volume:** Unlike ideal gases, real gas molecules occupy space. This reduces the volume available for gas movement, as noted with the constant \(b\).

These deviations are particularly noticeable in gases like \(\mathrm{CCl}_{4}\) which have significant Van der Waals constants \(a\) and \(b\). Factors that influence the degree of deviation include:

• **Temperature:** Lower temperatures enhance the effect of intermolecular attractions.
• **Pressure:** High pressures bring molecules closer, amplifying volume exclusions and interaction forces.

Understanding these deviations allows us to predict when the ideal gas law may not provide accurate results and when the Van der Waals equation is necessary.