Answer :
To calculate the volume of O₂(g) in a cylinder at 265 K and 1.00 atm, we can use the Ideal Gas Law formula:
[tex]\[ PV = nRT \][/tex]
Let's break down the steps and components needed:
1. Identify the variables:
- [tex]\( P \)[/tex] (Pressure) = 1.00 atm
- [tex]\( T \)[/tex] (Temperature) = 265 K
- [tex]\( R \)[/tex] (Ideal Gas Constant) = 0.0821 atm·L/(mol·K)
- [tex]\( n \)[/tex] (Number of moles of O₂): This value needs to be provided or assumed for a complete calculation.
2. Assume or provide the number of moles (n):
- For this setup, assume [tex]\( n = 1 \)[/tex] mole for simplicity.
3. Rearrange the Ideal Gas Law to solve for volume [tex]\( V \)[/tex]:
[tex]\[ V = \frac{nRT}{P} \][/tex]
4. Substitute the given values into the equation:
[tex]\[ V = \frac{(1 \, \text{mol}) \times (0.0821 \, \text{atm·L/(mol·K)}) \times (265 \, \text{K})}{1.00 \, \text{atm}} \][/tex]
5. Perform the calculation:
[tex]\[ V = \frac{(1) \times (0.0821) \times (265)}{1.00} \][/tex]
[tex]\[ V = 0.0821 \times 265 \][/tex]
[tex]\[ V = 21.7465 \][/tex]
Therefore, the volume [tex]\( V \)[/tex] of O₂(g) in the cylinder is approximately [tex]\( 21.75 \, \text{liters} \)[/tex], assuming 1 mole of gas.
If the actual number of moles [tex]\( n \)[/tex] of O₂ is different, you should substitute the correct value into the calculation.
[tex]\[ PV = nRT \][/tex]
Let's break down the steps and components needed:
1. Identify the variables:
- [tex]\( P \)[/tex] (Pressure) = 1.00 atm
- [tex]\( T \)[/tex] (Temperature) = 265 K
- [tex]\( R \)[/tex] (Ideal Gas Constant) = 0.0821 atm·L/(mol·K)
- [tex]\( n \)[/tex] (Number of moles of O₂): This value needs to be provided or assumed for a complete calculation.
2. Assume or provide the number of moles (n):
- For this setup, assume [tex]\( n = 1 \)[/tex] mole for simplicity.
3. Rearrange the Ideal Gas Law to solve for volume [tex]\( V \)[/tex]:
[tex]\[ V = \frac{nRT}{P} \][/tex]
4. Substitute the given values into the equation:
[tex]\[ V = \frac{(1 \, \text{mol}) \times (0.0821 \, \text{atm·L/(mol·K)}) \times (265 \, \text{K})}{1.00 \, \text{atm}} \][/tex]
5. Perform the calculation:
[tex]\[ V = \frac{(1) \times (0.0821) \times (265)}{1.00} \][/tex]
[tex]\[ V = 0.0821 \times 265 \][/tex]
[tex]\[ V = 21.7465 \][/tex]
Therefore, the volume [tex]\( V \)[/tex] of O₂(g) in the cylinder is approximately [tex]\( 21.75 \, \text{liters} \)[/tex], assuming 1 mole of gas.
If the actual number of moles [tex]\( n \)[/tex] of O₂ is different, you should substitute the correct value into the calculation.
