Problem 33 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

Calculate the temperature at which 20.0 mol of helium would exert a pressure of 120 atm in a \(10.0 \mathrm{dm}^{3}\) cylinder, using (a) the ideal gas equation and (b) the van der Waals equation. For He, \(a=0.034 \mathrm{dm}^{6}\) atm \(\mathrm{mol}^{-2}\) and \(b=0.024 \mathrm{dm}^{3} \mathrm{mol}^{-1}\). (Section 8.6)

Step-by-Step Solution

Answer

Using the ideal gas law, \( T \approx 730.65 \text{ K} \); using van der Waals, \( T \approx 701.54 \text{ K} \).

Step 1: Understand the Ideal Gas Law

The ideal gas equation is given by \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant (0.0821 dm³ atm mol⁻¹ K⁻¹), and \( T \) is the temperature in Kelvin.

Step 2: Rearrange the Ideal Gas Equation for Temperature

Solve for \( T \) by rearranging the ideal gas equation: \( T = \frac{PV}{nR} \).

Step 3: Substitute Values into the Ideal Gas Equation

Substitute \( P = 120 \) atm, \( V = 10.0 \) dm³, \( n = 20.0 \) mol, and \( R = 0.0821 \) dm³ atm mol⁻¹ K⁻¹ into the equation: \[ T = \frac{120 \times 10.0}{20.0 \times 0.0821} \].

Step 4: Calculate Temperature Using Ideal Gas Law

Calculate the value from the substitution: \[ T = \frac{1200}{1.642} \approx 730.65 \text{ K} \].

Step 5: Understand Van der Waals Equation

The Van der Waals equation accounts for real gas behavior and is given by: \[ \left( P + \frac{an^2}{V^2} \right)(V - nb) = nRT \].

Step 6: Rearrange Van der Waals Equation for Temperature

Solve for \( T \): \[ T = \frac{\left( P + \frac{an^2}{V^2} \right)(V - nb)}{nR} \].

Step 7: Substitute Values into van der Waals Equation

Substitute \( P = 120 \) atm, \( V = 10.0 \) dm³, \( n = 20.0 \) mol, \( a = 0.034 \) dm⁶ atm mol⁻², \( b = 0.024 \) dm³ mol⁻¹, and \( R = 0.0821 \) dm³ atm mol⁻¹ K⁻¹ into the equation: \[ T = \frac{\left( 120 + \frac{0.034 \cdot 20^2}{10^2} \right)(10 - 20 \cdot 0.024)}{20 \cdot 0.0821} \].

Step 8: Calculate Temperature Using van der Waals Equation

Calculate the intermediate terms: 1. \( \frac{0.034 \cdot 20^2}{10^2} = 1.36 \).2. \( 10 - 20 \times 0.024 = 9.52 \).Now substitute these back into the equation for \( T \): \[ T = \frac{(120 + 1.36)(9.52)}{1.642} = \frac{1151.0272}{1.642} \approx 701.54 \text{ K} \].

Key Concepts

Van der Waals Equation

When we talk about real gases, we dive into the Van der Waals equation, which provides a more accurate description than the Ideal Gas Law. This equation takes into consideration two key factors that affect gas behavior in real-life situations:

• The volume occupied by gas molecules themselves, captured by the "b" constant.
• The intermolecular forces between particles, which is represented by the "a" constant.

The Van der Waals equation is expressed as:\[\left( P + \frac{an^2}{V^2} \right)(V - nb) = nRT\]Here,

• \(P\): Pressure of the gas.
• \(V\): Volume available to the gas molecules.
• \(T\): Temperature in Kelvin.

By using this equation, which corrects for molecular interactions and volume, we get a more precise calculation of gas behavior under non-ideal conditions.

Real Gas Behavior

In real-life scenarios, gases don't always behave ideally. Real gas behavior diverges from the Ideal Gas Law due to intermolecular attractions and the finite size of gas molecules.
These deviations are more pronounced:

• At high pressures, where molecules are forced closer together.
• At low temperatures, where molecular motion slows, enhancing attractive forces.

Van der Waals' parameters, "a" and "b," help model these deviations. The constant "a" accounts for attractive forces, while "b" corrects for the volume occupied by gas particles. By applying these corrections, the Van der Waals equation offers an improved understanding of gas behavior under these challenging conditions.

Temperature Calculation

Temperature calculations are essential in understanding gas behavior. When using gas laws, we often solve for temperature to predict how gases will react under different conditions.
First, we use the Ideal Gas Law:\[ PV = nRT\]Here, solving for temperature \(T\) gives us:\[ T = \frac{PV}{nR}\]For real gases, the calculation becomes more complex as we turn to the Van der Waals equation:\[ T = \frac{(P + \frac{an^2}{V^2})(V - nb)}{nR}\]Understanding these equations and substitutions ensures accurate temperature determination, pivotal for industrial applications and scientific experiments alike.

Gas Laws

Gas laws form the foundation of our understanding of gas behavior, providing critical insight into how gases respond to changes in pressure, volume, and temperature.

• The **Ideal Gas Law**, \(PV = nRT\), gives a simplified model for gas behavior, assuming no interactions between molecules.
• **Boyle's Law**, for instance, describes how pressure and volume are inversely related at constant temperature.
• **Charles's Law** establishes the direct relationship between volume and temperature at constant pressure.

In real-world applications, gases often deviate from ideal conditions, prompting the use of amendments like the Van der Waals equation. These tailored laws help engineers, scientists, and students predict gas behaviors in a variety of contexts, from atmospheric studies to chemical reactions.