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 115 .

Short answer

The pressure exerted by the N₂ gas using the ideal gas law is \(12.3~atm\). Using the van der Waals equation, the pressure is also approximately \(12.3~atm\). The results from both methods are almost identical, showing that the ideal gas law is a good approximation for this case. Comparing these results with Exercise 115, where the pressure was also calculated to be \(12.3~atm\), suggests that all three methods are appropriate for N₂ gas under the given conditions, with a minor difference of around \(0.0017~atm\).

Step-by-step solution

a. Using the ideal gas law

First, we need to convert the given temperature from Celsius to Kelvin: \(T_K = T_C + 273.15\) For a temperature of \(25.0^\circ C\), in Kelvin it would be: \(T_K = 25.0 + 273.15 = 298.15K\) Now, we can plug in the given values into the ideal gas law equation, and solve for the pressure (P): \(PV = nRT \Rightarrow P = \frac{nRT}{V}\) Plugging in the values: \(n = 0.5000~mol\), \(V = 10.000~L\), \(R = 0.08206\frac{L\cdot atm}{mol\cdot K}\), \(T = 298.15~K\) \[P = \frac{(0.5000)(0.08206)(298.15)}{10.000}\] Calculate the pressure: \[P = 12.3~atm\] The pressure exerted by the N₂ gas using the ideal gas law is \(12.3~atm\).

b. Using the van der Waals equation

To use the van der Waals equation, we need the values of 'a' and 'b' for N₂ gas. Given that \(a = 1.39\frac{L^2\cdot atm}{mol^2}\) and \(b = 0.0391\frac{L}{mol}\), we can plug these values into the van der Waals equation and solve for the pressure (P). \(\left(P + \frac{n^2a}{V^2} \right) \left(V - nb\right) = nRT \) First, substitute the given values and constants into the equation: \(\left(P + \frac{(0.5000)^2(1.39)}{(10.000)^2} \right) \left(10.000 - (0.5000)(0.0391)\right) = (0.5000)(0.08206)(298.15)\) Now, simplify and solve for P: \(\left(P + 0.00173625\right)(9.98045) = 12.24756\) Multiply both sides by \(\frac{1}{9.98045}\), then subtract \(0.00173625\) to find the pressure: \[P = 12.3 - 0.0017\] Calculate the pressure: \[P = 12.3~atm\] The pressure exerted by the N₂ gas using the van der Waals equation is approximately \(12.3~atm\).

c. Compare the results

The results obtained from both the ideal gas law and van der Waals equation are almost identical, with a minor difference of around \(0.0017~atm\). This shows that for this particular case, the ideal gas law is a good approximation of the behavior of the N₂ gas.

d. Compare the results with those in Exercise 115

In Exercise 115, we calculated the pressure of N₂ gas using the pressure correction factor in the real gas equation, and the result was \(12.3~atm\). Comparing this result with the values obtained in parts a and b of this exercise, we can see that all three methods yield very similar results for the pressure exerted by the N₂ gas, with a difference of around \(0.0017~atm\). This suggests that all three methods are appropriate for N₂ gas under the given conditions.

Key concepts

Van der Waals Equation

The Van der Waals equation improves upon the Ideal Gas Law by accounting for the finite size of molecules and the intermolecular forces present in real gases. While the Ideal Gas Law presumes that gas particles have no volume and do not attract or repel each other, real gases clearly do not follow this assumption. The equation is expressed as:
\[\begin{equation}\left(P + \frac{n^2a}{V^2} \right) \left(V - nb\right) = nRT\end{equation}\]
where:

• \(P\) is the pressure of the gas,
• \(n\) is the number of moles of the gas,
• \(V\) is the volume of the gas,
• \(T\) is the temperature in Kelvin,
• \(R\) is the universal gas constant,
• \(a\) and \(b\) are Van der Waals constants that are specific to each gas, accounting for the gas particle's size and the strength of intermolecular forces, respectively.

In practical use, the Van der Waals equation can provide a more accurate description of gas behavior, particularly under conditions of high pressure and low temperature where deviations from ideal behavior are most pronounced.

Gas Pressure Calculation

The measure of force exerted by gas particles as they collide with the walls of a container is known as gas pressure. This fundamental concept is crucial in thermodynamics and can be calculated using several different equations, depending on the assumptions made about the gas.
For an ideal gas, the equation is simple, as shown in the Ideal Gas Law:
\(P = \frac{nRT}{V}\)
This equation states that the pressure (P) of the gas is directly proportional to the number of moles (n), the gas constant (R), and the temperature (T), and inversely proportional to the volume (V).
However, as seen with the Van der Waals equation, for real gases these calculations are adjusted to factor in intermolecular forces and the volume occupied by the gas particles themselves. The ability to accurately calculate gas pressure is critical in various fields, from chemistry to engineering, and it allows us to predict how gases will behave under changing conditions.

Real Gas Behavior

Real gases display behaviors that deviate from the Ideal Gas Law at high pressures and low temperatures. These deviations occur due to the volume occupied by the gas molecules and the forces that they exert on each other. Under typical laboratory conditions, real gases often behave similarly to ideal gases, allowing scientists to use the simpler Ideal Gas Law for calculations.
However, when conditions are far from standard temperature and pressure, or when high precision is necessary, corrections like those provided by the Van der Waals equation are used. The real gas behavior considers factors such as non-zero particle volume and attractions between particles, which can significantly affect how the gas expands, compresses, and how much pressure it exerts. Understanding real gas behavior is essential not only in academic settings but also in industries that handle gases under a variety of pressures and temperatures, ensuring safe and efficient operations.