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 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.

Short answer

(a) The pressure of ${\mathrm{CCl}}_4$ exerted at ${40}^{\circ }\mathrm{C}$ assuming ideal-gas behavior is $77.5\mathrm{kPa}$. (b) The pressure of ${\mathrm{CCl}}_4$ exerted at ${40}^{\circ }\mathrm{C}$ assuming van der Waals equation is $75.2\mathrm{kPa}$. (c) ${\mathrm{CCl}}_4$ is expected to deviate more from ideal behavior under these conditions than ${\mathrm{Cl}}_2$ due to the larger van der Waals constants.

Step-by-step solution

Convert Temperature to Kelvin

First, we need to convert the temperature given in Celsius to Kelvin. The temperature conversion formula is: $$T(K)=T{(}^{\circ }C)+273.15$$ In this case, the temperature is given as ${40}^{\circ }\mathrm{C}$, so: $$T(K)=40+273.15=313.15\mathrm{K}$$

Use the Ideal Gas Equation to Calculate Pressure

Using the ideal gas equation, we can find the pressure exerted by ${\mathrm{CCl}}_4$. The ideal gas equation is as follows: $$PV=nRT$$ Where: P = pressure, V = volume, n = moles of gas, R = ideal gas constant (8.314 J/(mol·K)), T = temperature. We are solving for P: $$P=\frac{nRT}{V}$$ Given: n = 1.00 mol, R = 8.314 J/(mol·K), T = 313.15 K, V = 33.3 L. First, we have to convert L to m³ since R is in J/(mol·K). $$33.3\mathrm{L}=0.0333{\mathrm{m}}^3$$ Now, substitute the values: $$P=\frac{(1.00\mathrm{mol})(8.314\mathrm{J}/(\mathrm{mol}\cdot \mathrm{K}))(313.15\mathrm{K})}{0.0333{\mathrm{m}}^3}$$ $$P=77.5\mathrm{kPa}$$ (a) The pressure of ${\mathrm{CCl}}_4$ exerted at ${40}^{\circ }\mathrm{C}$ assuming ideal-gas behavior is $77.5\mathrm{kPa}$.

Calculate Pressure with the Van Der Waals Equation

Now, we will calculate the pressure using the van der Waals equation. The equation is given as: $$P=\frac{nRT}{V-nb}-\frac{a{n}^2}{V^2}$$ From Table 10.3, we can find the van der Waals constants for ${\mathrm{CCl}}_4$. These are: a = 20.39 J·m³/mol², b = 0.1381 L/mol. Substituting the values from step 1, we get: $$P=\frac{(1.00\mathrm{mol})(8.314\mathrm{J}/(\mathrm{mol}\cdot \mathrm{K}))(313.15\mathrm{K})}{(0.0333{\mathrm{m}}^3-(1\mathrm{mol})(0.0001381{\mathrm{m}}^3/\mathrm{mol}))}-\frac{(20.39\mathrm{J}\cdot {\mathrm{m}}^3/({\mathrm{mol}}^2))(1^2{\mathrm{mol}}^2)}{(0.0333{\mathrm{m}}^3{)}^2}$$ $$P=75.2\mathrm{kPa}$$ (b) The pressure of ${\mathrm{CCl}}_4$ exerted at ${40}^{\circ }\mathrm{C}$ assuming van der Waals equation is $75.2\mathrm{kPa}$.

Comparison of Ideal Behavior Deviation

To determine which gas would deviate more from ideal behavior under these conditions (${\mathrm{Cl}}_2$ or ${\mathrm{CCl}}_4$), we will inspect the van der Waals constants. Gases with larger values of "a" and "b" experience more significant intermolecular forces and particle volume, respectively, causing more significant deviations from ideal behavior. Using Table 10.3 again, the van der Waal constants for ${\mathrm{Cl}}_2$ are: a = 6.49 J·m³/mol², b = 0.0562 L/mol. Comparing these values to those for ${\mathrm{CCl}}_4$: A(${\mathrm{Cl}}_2$) < A(${\mathrm{CCl}}_4$), B(${\mathrm{Cl}}_2$) < B(${\mathrm{CCl}}_4$). (c) Therefore, ${\mathrm{CCl}}_4$ is expected to deviate more from ideal behavior under these conditions than ${\mathrm{Cl}}_2$ due to the larger van der Waals constants.

Key concepts

Ideal Gas Law

Understanding the ideal gas law is crucial when delving into the behavior of gases under various conditions. It is an equation of state for a hypothetical gas called an 'ideal gas', which provides a good approximation to real gas behavior at high temperature and low pressure.

The ideal gas law is represented by the formula: $$PV=nRT$$
where:

• $P$ denotes the pressure exerted by the gas,
• $V$ represents the volume occupied by the gas,
• $n$ is the number of moles of the gas,
• $R$ is the universal gas constant, which is equal to 8.314 J/(mol·K), and
• $T$ is the absolute temperature, measured in Kelvin.

Using the ideal gas law for pressure calculation provides a simplified model that assumes gases have no intermolecular forces and the molecules occupy no volume. This assumption suits many practical situations, although for precise calculations, especially at high pressures and low temperatures, real gas behavior must be considered through models like the van der Waals equation.

Pressure Calculation

Pressure calculation involves determining the force exerted by a gas per unit area. In the context of gases, this force is due to the constant motion and collisions of gas particles with the container walls. Pressure is a critical concept in numerous fields, from meteorology to engineering, and even to the study of biological systems.

Using the ideal gas law, the pressure can be calculated if the other variables (volume, temperature, and amount of gas) are known. It is important to convert temperatures to Kelvin and volumes to the correct units to match the gas constant used. The pressure calculation step in our exercise provides a clear example of how to apply the ideal gas law in a practical scenario and solve for the pressure exerted by a gas.

Intermolecular Forces

Intermolecular forces are the forces of attraction or repulsion between neighboring particles (atoms, molecules, or ions). These forces play a significant role in determining the physical properties of substances, such as boiling and melting points, vapor pressure, and viscosity.

In gases, intermolecular forces are typically weak, allowing gas molecules to move freely and occupy the entire container. However, these forces cannot be entirely ignored, especially when gases are subjected to high pressures or low temperatures. Under these conditions, gases deviate from ideal behavior as predicted by the ideal gas law, due to the increased significance of intermolecular forces and the finite volume occupied by molecules.

The van der Waals equation offers a way to account for these factors through its inclusion of correction terms for intermolecular forces and molecular volume. By comparing the van der Waals constants for different gases, we can infer the relative magnitude of their intermolecular forces and expectancies of deviating from ideal behavior, a comparison that becomes particularly insightful in the context of our exercise's step 4.