IDNLearn.com: Your destination for reliable and timely answers to any question. Ask any question and get a detailed, reliable answer from our community of experts.
Sagot :
To determine the moles of air contained in a 147-L cylinder at a pressure of 1.17 atm and a temperature of 341 K, we can use the ideal gas law, which is stated as:
[tex]\[ PV = nRT \][/tex]
where:
- [tex]\( P \)[/tex] is the pressure in atmospheres (atm)
- [tex]\( V \)[/tex] is the volume in liters (L)
- [tex]\( n \)[/tex] is the number of moles of the gas
- [tex]\( R \)[/tex] is the ideal gas constant ([tex]\(0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \)[/tex])
- [tex]\( T \)[/tex] is the temperature in Kelvin (K)
The goal is to solve for [tex]\( n \)[/tex], the number of moles.
Step-by-step solution:
1. Identify the given values:
- Volume [tex]\( V = 147 \, \text{L} \)[/tex]
- Pressure [tex]\( P = 1.17 \, \text{atm} \)[/tex]
- Temperature [tex]\( T = 341 \, \text{K} \)[/tex]
- Ideal gas constant [tex]\( R = 0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \)[/tex]
2. Rearrange the ideal gas law to solve for [tex]\( n \)[/tex]:
[tex]\[ n = \frac{PV}{RT} \][/tex]
3. Substitute the known values into the equation:
[tex]\[ n = \frac{(1.17 \, \text{atm}) \times (147 \, \text{L})}{(0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K}) \times (341 \, \text{K})} \][/tex]
4. Calculate the numerator:
[tex]\[ (1.17 \, \text{atm}) \times (147 \, \text{L}) = 171.99 \, \text{atm} \cdot \text{L} \][/tex]
5. Calculate the denominator:
[tex]\[ (0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K}) \times (341 \, \text{K}) = 27.9961 \, \text{L} \cdot \text{atm} / \text{mol} \][/tex]
6. Divide the numerator by the denominator to find [tex]\( n \)[/tex]:
[tex]\[ n = \frac{171.99 \, \text{atm} \cdot \text{L}}{27.9961 \, \text{L} \cdot \text{atm} / \text{mol}} \approx 6.14 \, \text{mol} \][/tex]
Therefore, the number of moles of air contained in the 147-L cylinder at a pressure of 1.17 atm and a temperature of 341 K is approximately [tex]\( 6.14 \)[/tex] moles.
[tex]\[ PV = nRT \][/tex]
where:
- [tex]\( P \)[/tex] is the pressure in atmospheres (atm)
- [tex]\( V \)[/tex] is the volume in liters (L)
- [tex]\( n \)[/tex] is the number of moles of the gas
- [tex]\( R \)[/tex] is the ideal gas constant ([tex]\(0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \)[/tex])
- [tex]\( T \)[/tex] is the temperature in Kelvin (K)
The goal is to solve for [tex]\( n \)[/tex], the number of moles.
Step-by-step solution:
1. Identify the given values:
- Volume [tex]\( V = 147 \, \text{L} \)[/tex]
- Pressure [tex]\( P = 1.17 \, \text{atm} \)[/tex]
- Temperature [tex]\( T = 341 \, \text{K} \)[/tex]
- Ideal gas constant [tex]\( R = 0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \)[/tex]
2. Rearrange the ideal gas law to solve for [tex]\( n \)[/tex]:
[tex]\[ n = \frac{PV}{RT} \][/tex]
3. Substitute the known values into the equation:
[tex]\[ n = \frac{(1.17 \, \text{atm}) \times (147 \, \text{L})}{(0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K}) \times (341 \, \text{K})} \][/tex]
4. Calculate the numerator:
[tex]\[ (1.17 \, \text{atm}) \times (147 \, \text{L}) = 171.99 \, \text{atm} \cdot \text{L} \][/tex]
5. Calculate the denominator:
[tex]\[ (0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K}) \times (341 \, \text{K}) = 27.9961 \, \text{L} \cdot \text{atm} / \text{mol} \][/tex]
6. Divide the numerator by the denominator to find [tex]\( n \)[/tex]:
[tex]\[ n = \frac{171.99 \, \text{atm} \cdot \text{L}}{27.9961 \, \text{L} \cdot \text{atm} / \text{mol}} \approx 6.14 \, \text{mol} \][/tex]
Therefore, the number of moles of air contained in the 147-L cylinder at a pressure of 1.17 atm and a temperature of 341 K is approximately [tex]\( 6.14 \)[/tex] moles.
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Find clear answers at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.