Chapter 11: Problem 76
What volume (in liters) is occupied by \(3.38 \times 10^{22}\) nitrogen molecules at \(100^{\circ} \mathrm{C}\) and \(255 \mathrm{~mm} \mathrm{Hg} ?\)
Short Answer
Expert verified
The volume occupied is approximately 5.11 liters.
Step by step solution
01
Convert Pressure to Atmospheres
Pressure is given in millimeters of mercury (mm Hg). Firstly, convert this to atmospheres as it's a standard unit in gas law calculations. Use the conversion factor: \[1 ext{ atm} = 760 ext{ mm Hg}\] Thus, \[P = \frac{255 ext{ mm Hg}}{760 ext{ mm Hg/atm}} = 0.3355 ext{ atm}\]
02
Convert Temperature to Kelvin
Temperature is given in degrees Celsius. Convert it to Kelvin by adding 273.15 to the Celsius temperature: \[T = 100^{\circ}C + 273.15 = 373.15 ext{ K}\]
03
Calculate Moles of Gas
Use Avogadro's number to convert molecules to moles: Avogadro's Number = \(6.022 \times 10^{23} \text{ molecules/mol}\).Number of moles \(n\) is calculated as: \[n = \frac{3.38 \times 10^{22} \text{ molecules}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 0.0561 \text{ mol}\]
04
Use the Ideal Gas Law
The ideal gas law is \[PV = nRT\]where \(R\) is the ideal gas constant, \(0.0821 ext{ L atm/mol K}\). Substitute the known values into the equation:\[V = \frac{nRT}{P} = \frac{0.0561 \text{ mol} \times 0.0821 \text{ L atm/mol K} \times 373.15 \text{ K}}{0.3355 \text{ atm}}\]
05
Solve for Volume
Calculate the volume using the values from the previous steps:\[V \approx \frac{0.0561 \times 0.0821 \times 373.15}{0.3355} \approx 5.11 \text{ liters}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Pressure Conversion
Pressure is a critical component when utilizing the ideal gas law, and it's often necessary to convert pressure units to ensure consistency. In many exercises, like in this case, pressure is provided in millimeters of mercury (mm Hg). However, the ideal gas law formula typically requires pressure in atmospheres (atm).
To convert mm Hg to atm, use the conversion factor:
To convert mm Hg to atm, use the conversion factor:
- 1 atm = 760 mm Hg
- \( P = \frac{255 \text{ mm Hg}}{760 \text{ mm Hg/atm}} \approx 0.3355 \text{ atm} \)
Temperature Conversion
In gas law calculations, temperature must always be expressed in Kelvin. Kelvin is the absolute temperature scale and ensures that calculations remain consistent across various applications.
To convert Celsius to Kelvin, you simply add 273.15 to the temperature in degrees Celsius. This is because the Kelvin scale starts at absolute zero, which is approximately -273.15°C. For instance:
To convert Celsius to Kelvin, you simply add 273.15 to the temperature in degrees Celsius. This is because the Kelvin scale starts at absolute zero, which is approximately -273.15°C. For instance:
- If the temperature is 100°C, then the conversion to Kelvin is: \( T = 100^{\circ}C + 273.15 = 373.15 \text{ K} \)
Moles Calculation
Calculating the moles of a substance is crucial when using the ideal gas law. It allows one to connect the amount of gas to its volume, pressure, and temperature.
Avogadro's number provides the link between the number of molecules and moles. Avogadro’s number is a scientific constant with a value of:
Avogadro's number provides the link between the number of molecules and moles. Avogadro’s number is a scientific constant with a value of:
- \( 6.022 \times 10^{23} \text{ molecules/mol} \)
- If you have \( 3.38 \times 10^{22} \text{ molecules} \), the number of moles \( n \) is: \( n = \frac{3.38 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.0561 \text{ mol} \)
Avogadro's Number
Avogadro's number is a fundamental constant in chemistry and plays a crucial role in molecular and molar conversions. It denotes the number of particles – be they atoms, molecules, or ions – in one mole of a substance.
Avogadro's number is:
Avogadro's number is:
- \( 6.022 \times 10^{23} \text{ particles/mol} \)