Answer :
To determine the number of moles of air in a given volume, pressure, and temperature, we can use the Ideal Gas Law. The Ideal Gas Law is expressed as:
[tex]\[ PV = nRT \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure,
- [tex]\( V \)[/tex] is the volume,
- [tex]\( n \)[/tex] is the number of moles,
- [tex]\( R \)[/tex] is the ideal gas constant, and
- [tex]\( T \)[/tex] is the temperature in Kelvin.
Given:
- [tex]\( P = 1.21 \, \text{atm} \)[/tex]
- [tex]\( V = 134 \, \text{L} \)[/tex]
- [tex]\( T = 309 \, \text{K} \)[/tex]
- [tex]\( R = 0.0821 \, \text{(L} \cdot \text{atm)} / (\text{K} \cdot \text{mol}) \)[/tex]
We need to solve for [tex]\( n \)[/tex] (the number of moles). Rearranging the Ideal Gas Law to solve for [tex]\( n \)[/tex], we get:
[tex]\[ n = \frac{PV}{RT} \][/tex]
Substitute the given values into the equation:
[tex]\[ n = \frac{(1.21 \, \text{atm}) \times (134 \, \text{L})}{(0.0821 \, \text{(L} \cdot \text{atm)} / (\text{K} \cdot \text{mol}) ) \times (309 \, \text{K})} \][/tex]
Performing these calculations yields:
[tex]\[ n \approx 6.39 \][/tex]
So, the number of moles of air contained in the 134-L cylinder at a pressure of 1.21 atm and a temperature of 309 K is approximately [tex]\( 6.39 \)[/tex] moles.
[tex]\[ PV = nRT \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure,
- [tex]\( V \)[/tex] is the volume,
- [tex]\( n \)[/tex] is the number of moles,
- [tex]\( R \)[/tex] is the ideal gas constant, and
- [tex]\( T \)[/tex] is the temperature in Kelvin.
Given:
- [tex]\( P = 1.21 \, \text{atm} \)[/tex]
- [tex]\( V = 134 \, \text{L} \)[/tex]
- [tex]\( T = 309 \, \text{K} \)[/tex]
- [tex]\( R = 0.0821 \, \text{(L} \cdot \text{atm)} / (\text{K} \cdot \text{mol}) \)[/tex]
We need to solve for [tex]\( n \)[/tex] (the number of moles). Rearranging the Ideal Gas Law to solve for [tex]\( n \)[/tex], we get:
[tex]\[ n = \frac{PV}{RT} \][/tex]
Substitute the given values into the equation:
[tex]\[ n = \frac{(1.21 \, \text{atm}) \times (134 \, \text{L})}{(0.0821 \, \text{(L} \cdot \text{atm)} / (\text{K} \cdot \text{mol}) ) \times (309 \, \text{K})} \][/tex]
Performing these calculations yields:
[tex]\[ n \approx 6.39 \][/tex]
So, the number of moles of air contained in the 134-L cylinder at a pressure of 1.21 atm and a temperature of 309 K is approximately [tex]\( 6.39 \)[/tex] moles.