Problem 96 - Chemistry: The Central Science

Question

Calculate the pressure that \(\mathrm{CCl}_{4}\) will exert at \(40^{\circ} \mathrm{C}\) if \(1.00 \mathrm{~mol}\) occupies \(33.3 \mathrm{~L}\), assuming that (a) \(\mathrm{CCl}_{4}\) obeys the idealgas 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

Under ideal gas conditions, CCl4 will exert a pressure of 2.46 atm at 40°C, and under van der Waals conditions, it will exert a pressure of 2.35 atm. CCl4 will deviate more from ideal behavior than Cl2 under these conditions due to its stronger intermolecular forces and larger molecular size.

Step 1: (a) Pressure using Ideal Gas Equation

To calculate the pressure using the ideal gas equation, we use the formula: \(P=\frac{nRT}{V}\), where: P = pressure n = number of moles of gas (\(1.00~mol\)) R = ideal gas constant (\(0.0821~\frac{L.atm}{mol.K}\)) T = temperature in Kelvin (40°C + 273.15 = 313.15 K) V = volume occupied by gas (\(33.3~L\)) Now, use these values to find the pressure exerted by CCl4: \(P=\frac{1.00 * 0.0821 * 313.15}{33.3} =2.46~atm\) Answer: Under ideal gas conditions, CCl4 will exert a pressure of 2.46 atm at 40°C.

Step 2: (b) Pressure using Van der Waals Equation

Now, we will calculate the pressure using the van der Waals equation. The equation is: \[\left(P+\frac{a}{V_m^2}\right)(V_m-b)=RT\] where: P = pressure a = van der Waals constant a (Table 10.3) b = van der Waals constant b (Table 10.3) \(V_m\) = molar volume (\(\frac{V}{n}\)) T = temperature in Kelvin (313.15 K) From Table 10.3, for CCl4: \(a = 20.39~\frac{L^2.atm}{mol^2}\) \(b = 0.1282~\frac{L}{mol}\) Calculate the molar volume: \(V_m = \frac{33.3~L}{1.00~mol} = 33.3~\frac{L}{mol}\) Now, substitute these values into the van der Waals equation, and solve for P: \[\left(P+\frac{20.39}{(33.3)^2}\right)(33.3-0.128)=0.0821*313.15\] Solving for P, we get: \(P=2.35~atm\) Answer: Under van der Waals conditions, CCl4 will exert a pressure of 2.35 atm at 40°C.

Step 3: (c) Deviation comparison between Cl2 and CCl4

To determine which gas will deviate more from ideal behavior, we need to compare their strengths of intermolecular attractions and the size of the gas molecules. A larger van der Waals constant a indicates stronger intermolecular forces, while a larger constant b indicates larger molecular size. From Table 10.3, for Cl2 and CCl4: Cl2: \(a=6.49~\frac{L^2.atm}{mol^2}\), \(b=0.0562~\frac{L}{mol}\) CCl4: \(a=20.39~\frac{L^2.atm}{mol^2}\), \(b=0.1282~\frac{L}{mol}\) We can see that CCl4 has a larger 'a' and 'b' constant than Cl2, which means that CCl4 has stronger intermolecular forces and larger molecular size. Due to these factors, CCl4 will deviate more from ideal behavior than Cl2 under the given conditions.

Key Concepts

van der Waals equation

The van der Waals equation is a crucial concept in understanding real gas behavior, as it accounts for the factors that the ideal gas law overlooks. Unlike the simple PV=nRT, the van der Waals equation \[\left(P+\frac{a}{V_m^2}\right)(V_m-b)=RT\]introduces two new constants, *a* and *b*. These constants are vital as they help adjust the pressure and volume calculations to reflect the actual behavior of gases.
- **Constant *a*:** It corrects for the intramolecular forces between gas molecules. The stronger these forces, the higher the value of *a*. These forces compress the gas, reducing the actual pressure exerted on the container's walls.
- **Constant *b*:** It accounts for the finite volume occupied by the gas molecules themselves. The larger the molecules, the higher the value of *b*.
These adjustments make the van der Waals equation more accurate in predicting the behavior of gases, especially under high pressure and low temperature, where deviations from ideal behavior are more pronounced.

intermolecular forces

Intermolecular forces play a significant role in the behavior of gases and are a central concept when using the van der Waals equation. They are the forces of attraction or repulsion that act between particles. In the context of gases, these forces can slightly pull molecules together, impacting their pressure behavior.
- **Types of forces:** Different types of forces include hydrogen bonding, dipole-dipole interactions, and London dispersion forces. Each type varies in strength and effect on gas behavior.
- **Influence on real gases:** In gases like \(\mathrm{CCl}_{4}\), stronger intermolecular forces lead to decreased pressure compared to the ideal prediction. This deviation is corrected by the van der Waals constant *a*, which represents these forces.
Understanding these forces is essential for predicting and manipulating the behavior of gases in various chemical and physical processes.

molecular size

Molecular size is another important factor influencing gas behavior. This concept helps explain why gases often deviate from ideal behavior predictions.
- **Physical space:** Larger molecules take up more physical space, reducing the free volume available in a container. This factor is addressed by the van der Waals constant *b*, which corrects the volume in the equation.
- **Impact on pressure:** When molecular sizes are significant, they can cause a gas to exert less or more pressure than expected. In the case of \(\mathrm{CCl}_{4}\), the relatively large molecular size compared to smaller molecules like \(\mathrm{Cl}_{2}\) results in a greater deviation from ideality.
Understanding molecular size helps in comprehending how gases behave in confined spaces and can influence calculations in practical applications such as chemical engineering and atmospheric studies.