Problem 96 - Chemistry The Molecular Science

Question

Calculate the densities of \(\mathrm{Cl}_{2}\) and of \(\mathrm{SO}_{2}\) at \(25^{\circ} \mathrm{C}\) and \(0.750 \mathrm{~atm} .\) Then, calculate the density of \(\mathrm{Cl}_{2}\) at \(35^{\circ} \mathrm{C}\) and \(0.750 \mathrm{~atm}\) and the density of \(\mathrm{SO}_{2}\) at \(25^{\circ} \mathrm{C}\) and \(2.60 \mathrm{~atm} .\)

Step-by-Step Solution

Answer

Densities are approximately: \( \mathrm{Cl}_{2} \) (25°C, 0.750 atm): 2.176 g/L; \( \mathrm{SO}_{2} \) (25°C, 0.750 atm): 2.013 g/L; \( \mathrm{Cl}_{2} \) (35°C, 0.750 atm): 2.105 g/L; \( \mathrm{SO}_{2} \) (25°C, 2.60 atm): 6.982 g/L.

Step 1: Understanding the Ideal Gas Law

The problem involves determining densities using the Ideal Gas Law, which is given by \( PV = nRT \). We will use the equation for density, \( \rho = \frac{PM}{RT} \), derived from this law, where \( P \) is the pressure, \( M \) is the molar mass of the gas, \( R \) is the ideal gas constant (0.0821 L·atm/mol·K), and \( T \) is the temperature in Kelvin.

Step 2: Calculate Molar Mass for each Gas

Determine the molar mass of \( \mathrm{Cl}_{2} \) and \( \mathrm{SO}_{2} \). For \( \mathrm{Cl}_{2} \), the molar mass is \( 2 \times 35.45 = 70.90 \) g/mol. For \( \mathrm{SO}_{2} \), the molar mass is \( 32.07 + 2\times 16.00 = 64.07 \) g/mol.

Step 3: Convert Temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin using \( T(K) = T(°C) + 273.15 \). For \( 25^{\circ} \mathrm{C} \), \( T = 298.15 \) K. For \( 35^{\circ} \mathrm{C} \), \( T = 308.15 \) K.

Step 4: Calculate Density of \( \mathrm{Cl}_{2} \) at 25°C and 0.750 atm

Apply the density formula: \( \rho = \frac{PM}{RT} = \frac{(0.750 \ ext{atm})(70.90 \ ext{g/mol})}{(0.0821 \ ext{L·atm/mol·K})(298.15 \ ext{K})} \approx 2.176 \ ext{g/L} \).

Step 5: Calculate Density of \( \mathrm{SO}_{2} \) at 25°C and 0.750 atm

Using the same formula: \( \rho = \frac{(0.750 \ ext{atm})(64.07 \ ext{g/mol})}{(0.0821 \ ext{L·atm/mol·K})(298.15 \ ext{K})} \approx 2.013 \ ext{g/L} \).

Step 6: Calculate Density of \( \mathrm{Cl}_{2} \) at 35°C and 0.750 atm

Using the density formula: \( \rho = \frac{(0.750 \ ext{atm})(70.90 \ ext{g/mol})}{(0.0821 \ ext{L·atm/mol·K})(308.15 \ ext{K})} \approx 2.105 \ ext{g/L} \).

Step 7: Calculate Density of \( \mathrm{SO}_{2} \) at 25°C and 2.60 atm

Using the density formula: \( \rho = \frac{(2.60 \ ext{atm})(64.07 \ ext{g/mol})}{(0.0821 \ ext{L·atm/mol·K})(298.15 \ ext{K})} \approx 6.982 \ ext{g/L} \).

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a crucial equation in chemistry, especially for calculations involving gases. It's expressed as \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the absolute temperature in Kelvin. This law helps us understand how gases will behave under different conditions.

• With our exercise, we're focusing on the density formula \( \rho = \frac{PM}{RT} \), which is derived from the Ideal Gas Law.
• In this context, \( M \) stands for molar mass, and \( \rho \) represents density.
• This relationship shows how pressure, molar mass, and temperature impact the density of a gas.

Being able to manipulate this equation is critical. The direct relationship between pressure and density and the inverse relationship between temperature and density explain how gases compress and expand.

Molar Mass

Molar mass is a fundamental concept that denotes the mass of a given substance (element or compound) per mole. It is expressed in grams per mole (g/mol). Understanding molar mass is essential when working with the Ideal Gas Law.

• For \( \mathrm{Cl}_{2} \), which is a diatomic molecule consisting of two chlorine atoms, the molar mass is determined as follows:
  • Chlorine's atomic mass is approximately 35.45. Doubling this number gives us a molar mass of 70.90 g/mol.
• For \( \mathrm{SO}_{2} \), composed of one sulfur and two oxygen atoms, the molar mass calculation is:
  • Sulfur's atomic mass is about 32.07, and oxygen's atomic mass is 16.00. So, we calculate: 32.07 + (2 \times 16.00) = 64.07 g/mol.

Acccurately calculating these values is key for determining gas densities using the Ideal Gas Law.

Temperature Conversion

Temperature plays a crucial role in calculations involving gases. Since the Ideal Gas Law requires temperature in Kelvin, it's important to properly convert temperatures that are typically given in degrees Celsius.

• The conversion formula is simple: \[ T(K) = T(°C) + 273.15 \]
• For instance, converting 25°C to Kelvin gives us 298.15 K.
• Similarly, 35°C converts to 308.15 K.

This conversion is imperative because Kelvin is an absolute scale, allowing for straightforward use in gas law equations where absolute zero (0 K) is a theoretical point where all kinetic energy stops.

Pressure and Temperature Conditions

The behaviors and properties of gases are significantly influenced by pressure and temperature conditions.

• Higher pressure tends to increase the density of a gas, as the gas particles are forced closer together.
• Conversely, raising the temperature at constant pressure decreases the density because gas particles move more vigorously and spread out more.
• In our exercise, we deal with varying conditions:
  • Density of gases like \( \mathrm{Cl}_{2} \) and \( \mathrm{SO}_{2} \) are calculated at pressures of 0.750 atm and 2.60 atm over different temperatures.

Clearly understanding the relationship between these conditions and gas density is vital for any situations where gas properties are of importance.