Chapter 11: Problem 19
A flask is first evacuated so that it contains no gas at all. Then, \(2.2 \mathrm{g}\) of \(\mathrm{CO}_{2}\) is introduced into the flask. On warming to \(22^{\circ} \mathrm{C}\), the gas exerts a pressure of \(318 \mathrm{mm}\) Hg. What is the volume of the flask?
Short Answer
Expert verified
The volume of the flask is approximately 1.22 liters.
Step by step solution
01
Understand the Problem
We need to find the volume of a flask containing 2.2 g of \( \mathrm{CO}_2 \). The gas exerts a pressure of 318 mmHg at a temperature of 22°C.
02
Convert Units Appropriately
Convert the pressure from mmHg to atm for consistency in calculation. \[ 1 \text{ atm} = 760 \text{ mmHg} \] thus, \( 318 \text{ mmHg} = \frac{318}{760} \text{ atm} \).
03
Calculate Moles of CO2
Determine the moles of the \( \mathrm{CO}_2 \). The molar mass of \( \mathrm{CO}_2 \) is 44.01 g/mol. Use \( n = \frac{mass}{molar mass} \) to get \( n = \frac{2.2}{44.01} \text{ mol} \).
04
Convert Temperature to Kelvin
Convert temperature from Celsius to Kelvin using the formula: \( T(K) = 22 + 273.15 \). This gives \( T = 295.15 \text{ K} \).
05
Apply the Ideal Gas Law
Use the Ideal Gas Law formula \( PV = nRT \) where \( R = 0.0821 \text{ L atm/mol K} \). Substitute the known values: \( P = \frac{318}{760} \text{ atm} \), \( n = \frac{2.2}{44.01} \text{ mol} \), \( T = 295.15 \text{ K} \). Solve for \( V \).
06
Calculate Volume
Rearrange the Ideal Gas Law to find \( V \): \[ V = \frac{nRT}{P} = \frac{\left(\frac{2.2}{44.01}\right) \times 0.0821 \times 295.15}{\left(\frac{318}{760}\right)} \] and compute \( V \) to find the volume of the flask.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mole Calculation
To calculate the moles of a gas, we use the relationship between mass, molar mass, and number of moles. In our exercise, we want to find the number of moles for carbon dioxide (CO\(_2\)) present, since knowing the amount of substance in moles is essential for using the Ideal Gas Law. Here are the simple steps:
- Identify the molar mass of CO\(_2\). Carbon dioxide, made up of one carbon and two oxygen atoms, has a molar mass of 44.01 g/mol.
- Use the formula to solve for moles: \( n = \frac{m}{M} \), where \( n \) is the number of moles, \( m \) is the mass of the gas in grams, and \( M \) is the molar mass.
Pressure Conversion
In many chemistry problems, converting pressure into a common unit is crucial for applying the Ideal Gas Law. Pressure is often measured in different units like mmHg or atm. Here, we'll convert pressure from mmHg to atm, to ensure we use consistent units throughout.
The exercise gives us a pressure of 318 mmHg. Here’s how to proceed to convert it into atmospheres (atm):
The exercise gives us a pressure of 318 mmHg. Here’s how to proceed to convert it into atmospheres (atm):
- Remember the conversion factor: 1 atm = 760 mmHg.
- Calculate the pressure in atm by dividing: \( 318 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}} \).
Temperature Conversion
When dealing with gas laws, converting temperature from Celsius to Kelvin is a straightforward but vital step. The Kelvin scale is used in scientific calculations to avoid negative temperatures, as Kelvin starts from absolute zero.
To convert from Celsius to Kelvin, you simply add 273.15 to the Celsius temperature:
To convert from Celsius to Kelvin, you simply add 273.15 to the Celsius temperature:
- Start with the Celsius temperature given in the problem, which is 22°C.
- Apply the conversion: \( T(\text{K}) = T(\degree \text{C}) + 273.15 \).
Volume Calculation
The Ideal Gas Law, expressed as \( PV = nRT \), is the key to calculating volume in this exercise. This formula relates the pressure \( P \), volume \( V \), and temperature \( T \) of a gas, along with the number of moles \( n \) and the ideal gas constant \( R \). Let's see how you can determine the volume:
- First, rearrange the Ideal Gas Law to solve for volume: \( V = \frac{nRT}{P} \).
- Plug in the values calculated earlier: \( n = \frac{2.2}{44.01} \) moles, \( T = 295.15 \; \text{K} \), and \( P = 0.418 \; \text{atm} \). Don't forget \( R = 0.0821 \; \text{L atm/mol K} \), the constant.