Chapter 10: Problem 49
Calculate the density of hydrogen bromide (HBr) gas in \(\mathrm{g} / \mathrm{L}\) at \(733 \mathrm{mmHg}\) and \(46^{\circ} \mathrm{C}\).
Short Answer
Expert verified
The density of HBr gas at the given conditions is approximately 2.98 g/L.
Step by step solution
01
Understanding the Problem
We need to calculate the density of hydrogen bromide (HBr) gas in grams per liter (g/L) at a pressure of 733 mmHg and a temperature of 46°C. To do this, we'll use the Ideal Gas Law and the concept of molar mass.
02
Convert Units
Before using the Ideal Gas Law, let's ensure all units are in the correct form. We need to convert 733 mmHg to atm and 46°C to Kelvin. To convert the temperature, add 273.15 to Celsius: \[ T(K) = 46 + 273.15 = 319.15 \text{ K} \]To convert the pressure, use the conversion factor 1 atm = 760 mmHg:\[ P(atm) = \frac{733}{760} \approx 0.9645 \text{ atm} \]
03
Ideal Gas Law Setup
The Ideal Gas Law 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 L.atm/mol.K), and \( T \) is the temperature in Kelvin.We can manipulate this to find the density formula: \[ \rho = \frac{PM}{RT} \] where \( \rho \) is density and \( M \) is the molar mass of the gas.
04
Molar Mass Calculation
Calculate the molar mass of HBr. Hydrogen (H) has an atomic mass of approximately 1.01 g/mol and Bromine (Br) has an atomic mass of approximately 79.90 g/mol. Therefore, the molar mass of HBr is:\[ M = 1.01 + 79.90 = 80.91 \text{ g/mol} \]
05
Calculate Density
Use the derived density formula \( \rho = \frac{PM}{RT} \) to find the density of HBr.Substitute the known values: \( P = 0.9645 \text{ atm} \), \( M = 80.91 \text{ g/mol} \), \( R = 0.0821 \text{ L.atm/mol.K} \), and \( T = 319.15 \text{ K} \).\[ \rho = \frac{0.9645 \times 80.91}{0.0821 \times 319.15} \approx \frac{78.08}{26.20} \approx 2.98 \text{ g/L} \]
06
Conclusion
The density of hydrogen bromide (HBr) gas at 733 mmHg and 46°C is approximately 2.98 g/L.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Density Calculations
When calculating the density of a gas, we're trying to find out how much mass is present in a certain volume. Density, usually symbolized as \( \rho \), is typically expressed in grams per liter (g/L). In the case of a gas, we can use the Ideal Gas Law to help us find this value if we also know the pressure, temperature, and the molar mass of the gas.
The formula derived from the Ideal Gas Law to calculate density is: \[ \rho = \frac{PM}{RT} \]where: - \( P \) is the pressure of the gas, - \( M \) is the molar mass of the gas, - \( R \) is the ideal gas constant (0.0821 L.atm/mol.K), - \( T \) is the temperature in Kelvin.
This formula allows us to calculate the density without directly measuring the mass and volume, making it very useful for gases.
The formula derived from the Ideal Gas Law to calculate density is: \[ \rho = \frac{PM}{RT} \]where: - \( P \) is the pressure of the gas, - \( M \) is the molar mass of the gas, - \( R \) is the ideal gas constant (0.0821 L.atm/mol.K), - \( T \) is the temperature in Kelvin.
This formula allows us to calculate the density without directly measuring the mass and volume, making it very useful for gases.
Molar Mass
The molar mass of a substance is a measure of the mass of one mole of that substance. It is expressed in grams per mole (g/mol). To calculate the molar mass of a compound, you need to add up the atomic masses of all the atoms in its chemical formula.
For hydrogen bromide (HBr), the molar mass is calculated as:- Hydrogen (H) has an atomic mass of approximately 1.01 g/mol- Bromine (Br) has an atomic mass of approximately 79.90 g/molThus, the molar mass of HBr is:\[ M = 1.01 + 79.90 = 80.91 \text{ g/mol} \] Understanding molar mass is crucial for density calculations, as it directly enters the formula to find density using the Ideal Gas Law.
For hydrogen bromide (HBr), the molar mass is calculated as:- Hydrogen (H) has an atomic mass of approximately 1.01 g/mol- Bromine (Br) has an atomic mass of approximately 79.90 g/molThus, the molar mass of HBr is:\[ M = 1.01 + 79.90 = 80.91 \text{ g/mol} \] Understanding molar mass is crucial for density calculations, as it directly enters the formula to find density using the Ideal Gas Law.
Unit Conversion
Using consistent units is key when performing calculations involving physical quantities. For gases, it often involves converting pressure from mmHg (millimeters of mercury) to atm (atmospheres) and temperature from Celsius to Kelvin.
Here’s how you can make these conversions:- To convert Celsius to Kelvin, add 273.15. So, for 46°C: \[ T(K) = 46 + 273.15 = 319.15 \text{ K} \]- To convert mmHg to atm, use the equivalence 1 atm = 760 mmHg. Hence, for 733 mmHg: \[ P(atm) = \frac{733}{760} \approx 0.9645 \text{ atm} \]Unit conversions ensure that all parts of the Ideal Gas Law equation are consistent and allow accurate calculations in scientific contexts.
Here’s how you can make these conversions:- To convert Celsius to Kelvin, add 273.15. So, for 46°C: \[ T(K) = 46 + 273.15 = 319.15 \text{ K} \]- To convert mmHg to atm, use the equivalence 1 atm = 760 mmHg. Hence, for 733 mmHg: \[ P(atm) = \frac{733}{760} \approx 0.9645 \text{ atm} \]Unit conversions ensure that all parts of the Ideal Gas Law equation are consistent and allow accurate calculations in scientific contexts.
Gas Laws
Gas laws are a set of rules that describe how gases behave in relation to the pressure, volume, and temperature they experience. The Ideal Gas Law, one of the most important of these laws, is an equation of state that helps relate all these factors into a single formula.The Ideal Gas Law 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,- \( T \) is the temperature in Kelvin.By manipulating this standard formula, we're able to derive expressions like the one for density, making it a versatile tool in chemistry and physics for studying gas behavior under various conditions.