IDNLearn.com provides a user-friendly platform for finding answers to your questions. Join our Q&A platform to get accurate and thorough answers to all your pressing questions.
Sagot :
To determine the number of moles of air inside the volleyball, we will use the ideal gas law, which is given by the formula:
[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 universal gas constant,
- [tex]\( T \)[/tex] is the temperature in Kelvin.
First, let's identify the given values and convert them to the proper units:
1. Pressure ([tex]\( P \)[/tex]):
- Given pressure: 130.75 kilopascals (kPa)
- Conversion to atmospheres (atm): [tex]\( 1 \text{ atm} = 101.3 \text{ kPa} \)[/tex]
- Therefore, pressure in atmospheres: [tex]\( \frac{130.75 \text{ kPa}}{101.3 \text{ kPa/atm}} \)[/tex]
2. Temperature ([tex]\( T \)[/tex]):
- Given temperature: [tex]\( 24.6^{\circ} \text{C} \)[/tex]
- Conversion to Kelvin: [tex]\( T(K) = 24.6 + 273.15 \)[/tex]
3. Volume ([tex]\( V \)[/tex]):
- Given volume: 5.27 liters (no conversion needed)
4. Universal gas constant ([tex]\( R \)[/tex]):
- Given as [tex]\( 0.0821 \frac{L \cdot \text{atm}}{\text{mol} \cdot K} \)[/tex]
To find the number of moles ([tex]\( n \)[/tex]), we rearrange the ideal gas law to solve for [tex]\( n \)[/tex]:
[tex]\[ n = \frac{PV}{RT} \][/tex]
Now, substitute the given values into the equation:
1. Convert pressure to atmospheres:
[tex]\[ P = \frac{130.75 \text{ kPa}}{101.3 \text{ kPa/atm}} \approx 1.290 \text{ atm} \][/tex]
2. Convert temperature to Kelvin:
[tex]\[ T = 24.6^{\circ} \text{C} + 273.15 = 297.75 \text{ K} \][/tex]
3. Using the given volume and gas constant:
[tex]\[ V = 5.27 \text{ L} \][/tex]
[tex]\[ R = 0.0821 \frac{L \cdot \text{atm}}{\text{mol} \cdot K} \][/tex]
Now, calculate the number of moles:
[tex]\[ n = \frac{(1.290 \text{ atm}) \times (5.27 \text{ L})}{(0.0821 \frac{L \cdot \text{atm}}{\text{mol} \cdot K}) \times (297.75 \text{ K})} \approx 0.278 \text{ mol} \][/tex]
Given the answer choices, the correct answer is:
A. 0.278 mol
[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 universal gas constant,
- [tex]\( T \)[/tex] is the temperature in Kelvin.
First, let's identify the given values and convert them to the proper units:
1. Pressure ([tex]\( P \)[/tex]):
- Given pressure: 130.75 kilopascals (kPa)
- Conversion to atmospheres (atm): [tex]\( 1 \text{ atm} = 101.3 \text{ kPa} \)[/tex]
- Therefore, pressure in atmospheres: [tex]\( \frac{130.75 \text{ kPa}}{101.3 \text{ kPa/atm}} \)[/tex]
2. Temperature ([tex]\( T \)[/tex]):
- Given temperature: [tex]\( 24.6^{\circ} \text{C} \)[/tex]
- Conversion to Kelvin: [tex]\( T(K) = 24.6 + 273.15 \)[/tex]
3. Volume ([tex]\( V \)[/tex]):
- Given volume: 5.27 liters (no conversion needed)
4. Universal gas constant ([tex]\( R \)[/tex]):
- Given as [tex]\( 0.0821 \frac{L \cdot \text{atm}}{\text{mol} \cdot K} \)[/tex]
To find the number of moles ([tex]\( n \)[/tex]), we rearrange the ideal gas law to solve for [tex]\( n \)[/tex]:
[tex]\[ n = \frac{PV}{RT} \][/tex]
Now, substitute the given values into the equation:
1. Convert pressure to atmospheres:
[tex]\[ P = \frac{130.75 \text{ kPa}}{101.3 \text{ kPa/atm}} \approx 1.290 \text{ atm} \][/tex]
2. Convert temperature to Kelvin:
[tex]\[ T = 24.6^{\circ} \text{C} + 273.15 = 297.75 \text{ K} \][/tex]
3. Using the given volume and gas constant:
[tex]\[ V = 5.27 \text{ L} \][/tex]
[tex]\[ R = 0.0821 \frac{L \cdot \text{atm}}{\text{mol} \cdot K} \][/tex]
Now, calculate the number of moles:
[tex]\[ n = \frac{(1.290 \text{ atm}) \times (5.27 \text{ L})}{(0.0821 \frac{L \cdot \text{atm}}{\text{mol} \cdot K}) \times (297.75 \text{ K})} \approx 0.278 \text{ mol} \][/tex]
Given the answer choices, the correct answer is:
A. 0.278 mol
Your presence in our community is highly appreciated. Keep sharing your insights and solutions. Together, we can build a rich and valuable knowledge resource for everyone. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.
