Question

Calculate the pressure exerted by 0.5000 mole of \(\mathrm{N}_{2}\) in a 10.000 -L container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results. d. Compare the results with those in Exercise 123.

Short answer

a. Using the Ideal Gas Law, the pressure exerted by the nitrogen gas is \(1.2217\: atm\). b. Using the van der Waals equation, the pressure is calculated as \(1.2244\: atm\). c. The difference between the two values is small (0.0027 atm), suggesting the Ideal Gas Law offers a good approximation in this case. d. Compared to Exercise 123, the pressure difference in this problem is less significant, indicating that the Ideal Gas Law is less accurate under the conditions of Exercise 123.

Step-by-step solution

Calculate the temperature in Kelvin

To calculate the pressure, we first need to convert the given temperature from Celsius to Kelvin. To do this, simply add 273.15 to the Celsius value. \( T_K = 25.0 + 273.15 = 298.15 \: K \)

a. Calculate the pressure using the Ideal Gas Law

The Ideal Gas Law is given by the equation: \(PV = nRT\) Where P = pressure, V = volume, n = number of moles, R = gas constant, and T = temperature in Kelvin. We are given n, V, and T and want to find P. The gas constant R for pressure in atm and volume in liters is 0.0821 L atm/mol K. Therefore, \( P = \frac{nRT}{V} \) Plug in the values, n = 0.5000 mol, R = 0.0821 L atm/mol K, T = 298.15 K, and V = 10.000 L: \( P = \frac{(0.5000)(0.0821)(298.15)}{10} = 1.2217\: atm\)

b. Calculate the pressure using the van der Waals equation

The van der Waals equation takes into account the molecular size and intermolecular forces. It is given by: \( \left(P + \frac{an^2}{V^2}\right)(V-nb) = nRT\) For N₂, van der Waals coefficients are \(a = 1.390\:L^2\: atm/mol^2\) and \(b = 0.0391\: L/mol^2\). We need to solve for P. To do this, let's isolate P in the equation: \( P = \frac{nRT}{V-nb} - \frac{an^2}{V^2}\) Now plug in the values, n = 0.5000 mol, R = 0.0821 L atm/mol K, T = 298.15 K, V = 10.000 L, a = 1.390 L² atm/mol², and b = 0.0391 L/mol: \( P = \frac{(0.5000)(0.0821)(298.15)}{10 - (0.5000)(0.0391)} - \frac{1.390(0.5000)^2}{10^2}\) \( P = 1.2244 \: atm \)

c. Compare the results

The pressures calculated using the Ideal Gas Law and the van der Waals equation are: Ideal Gas Law: \(P = 1.2217 \: atm\) van der Waals equation: \(P = 1.2244 \: atm\) The difference between these two values is small, only about 0.0027 atm. This suggests that the ideal gas law gives a good approximation of the gas pressure in this case.

d. Compare with Exercise 123

In exercise 123, we found that the pressure difference between the ideal gas law and the van der Waals equation was much more significant. This indicates that the ideal gas law is less accurate under those conditions than it is under the conditions of this problem.

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry for understanding how gases behave. It is expressed as \(PV = nRT\), where \(P\) represents pressure, \(V\) is volume, \(n\) is the number of moles of the gas, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. This equation assumes that gas molecules do not attract or repel each other and have no volume, which is why it’s labeled as "ideal." However, this assumption does not always hold true in real-world conditions.

When using the ideal gas law to calculate gas pressure, you only need to rearrange the formula to solve for \(P\). The simplicity of the ideal gas law makes it quite useful for quick calculations and approximations in everyday scenarios. However, it tends to be more accurate at high temperatures and low pressures. In cases where molecular interactions or the actual volume of gas particles cannot be ignored, other equations come into play.

Van der Waals Equation

The van der Waals equation provides a more realistic way to calculate pressure for a gas by taking into account the size of gas molecules and intermolecular forces. It modifies the ideal gas equation to include these factors. The formula is: \[ \left(P + \frac{an^2}{V^2}\right)(V-nb) = nRT \]
Here, \(a\) and \(b\) are unique coefficients for each gas, representing the strength of intermolecular attractions and the volume occupied by gas particles, respectively. The term \(\frac{an^2}{V^2}\) corrects for attraction between molecules, while \(nb\) corrects for the actual volume occupied.

Though it’s slightly more complex than the ideal gas law, using the van der Waals equation can give more accurate predictions of gas behavior under varying conditions, such as high pressure or low temperature, where the ideal gas assumptions about negligible size and no intermolecular forces are inadequate. You may notice small differences in calculated pressures between this and the ideal gas law.

Gas Pressure Calculation

Calculating gas pressure involves using theoretical equations to predict how gas molecules behave in a confined space. In this context, pressure is the result of gas molecules colliding with the walls of their container. There are two main ways to calculate this:

• Using the ideal gas law, where assumptions about molecular size and interactions are ignored. This method often results in a straightforward calculation: \(P = \frac{nRT}{V}\).
• Applying the van der Waals equation, which offers a refined calculation by considering molecular interactions and size: \(P = \frac{nRT}{V-nb} - \frac{an^2}{V^2}\).

While the ideal gas law is quicker and easier to compute, the van der Waals equation provides accurate insight, particularly when dealing with gases at high pressures or low temperatures. Each method gives slightly different values for pressure, reflecting how closely gas molecules follow the ideal assumptions.

Molecular Interactions

Molecules in a gas can interact in multiple ways, primarily dictated by their proximity to one another and the forces acting between them. These interactions become significant under certain conditions and can affect physical properties like pressure.

• Intermolecular Forces: These are attractions or repulsions between molecules. Van der Waals forces are weak attractions that exist even in "ideal" gases but become notable at high pressures and low temperatures.
• Molecular Size: In the ideal gas law, molecules are assumed to have no volume. However, in reality, they occupy space and this can impact calculations of gas properties.

When performing calculations using the van der Waals equation, you adjust for these factors, allowing a more comprehensive understanding of how gases behave beyond the idealized scenarios. These adjustments account for the non-ideal behavior exhibited by real gases, especially under extreme conditions.