Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
To determine the number of moles of oxygen (O₂) contained in a 3.6 L cylinder with a pressure of 2601 mmHg and a temperature of 31°C, we can use the Ideal Gas Law equation, \( PV = nRT \), where:
- \( P \) is pressure
- \( V \) is volume
- \( n \) is the number of moles
- \( R \) is the universal gas constant
- \( T \) is temperature in Kelvin
Here’s a step-by-step solution:
Step 1: Convert Temperature to Kelvin
We are given the temperature in Celsius (31°C), which we need to convert to Kelvin. The conversion formula is:
[tex]\[ T(K) = T(°C) + 273.15 \][/tex]
So,
[tex]\[ T(K) = 31 + 273.15 = 304.15 \, K \][/tex]
Step 2: Convert Pressure to atm
We are given the pressure in mmHg (2601 mmHg), which we need to convert to atmospheres (atm). The conversion factor is 1 atm = 760 mmHg. The conversion formula is:
[tex]\[ P(\text{atm}) = \frac{P(\text{mmHg})}{760} \][/tex]
So,
[tex]\[ P(\text{atm}) = \frac{2601}{760} = 3.422368421 \, atm \][/tex]
Step 3: Use the Ideal Gas Law to Solve for n (number of moles)
[tex]\[ PV = nRT \][/tex]
We can rearrange the equation to solve for \( n \):
[tex]\[ n = \frac{PV}{RT} \][/tex]
Substitute the values:
- \( P = 3.422368421 \, atm \)
- \( V = 3.6 \, L \)
- \( R = 0.0821 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}) \)
- \( T = 304.15 \, K \)
[tex]\[ n = \frac{(3.422368421 \, atm \times 3.6 \, L)}{(0.0821 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}) \times 304.15 \, K)} \][/tex]
[tex]\[ n = \frac{12.3205263156 \, \text{L} \cdot \text{atm}}{24.962215 \, \text{L} \cdot \text{atm/mole}} \][/tex]
[tex]\[ n = 0.4933990202 \, \text{moles} \][/tex]
Step 4: Round to Appropriate Significant Figures
Given the significant figures in the problem's measurements (2601 mmHg with four significant digits, 3.6 L with two significant digits, and 31°C with two significant digits), the final answer should be rounded to two significant digits.
Therefore, the number of moles of oxygen is:
[tex]\[ n ≈ 0.49 \, \text{moles} \][/tex]
So, the number of moles of oxygen (Oâ‚‚) contained in the cylinder is approximately 0.49 moles.
- \( P \) is pressure
- \( V \) is volume
- \( n \) is the number of moles
- \( R \) is the universal gas constant
- \( T \) is temperature in Kelvin
Here’s a step-by-step solution:
Step 1: Convert Temperature to Kelvin
We are given the temperature in Celsius (31°C), which we need to convert to Kelvin. The conversion formula is:
[tex]\[ T(K) = T(°C) + 273.15 \][/tex]
So,
[tex]\[ T(K) = 31 + 273.15 = 304.15 \, K \][/tex]
Step 2: Convert Pressure to atm
We are given the pressure in mmHg (2601 mmHg), which we need to convert to atmospheres (atm). The conversion factor is 1 atm = 760 mmHg. The conversion formula is:
[tex]\[ P(\text{atm}) = \frac{P(\text{mmHg})}{760} \][/tex]
So,
[tex]\[ P(\text{atm}) = \frac{2601}{760} = 3.422368421 \, atm \][/tex]
Step 3: Use the Ideal Gas Law to Solve for n (number of moles)
[tex]\[ PV = nRT \][/tex]
We can rearrange the equation to solve for \( n \):
[tex]\[ n = \frac{PV}{RT} \][/tex]
Substitute the values:
- \( P = 3.422368421 \, atm \)
- \( V = 3.6 \, L \)
- \( R = 0.0821 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}) \)
- \( T = 304.15 \, K \)
[tex]\[ n = \frac{(3.422368421 \, atm \times 3.6 \, L)}{(0.0821 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}) \times 304.15 \, K)} \][/tex]
[tex]\[ n = \frac{12.3205263156 \, \text{L} \cdot \text{atm}}{24.962215 \, \text{L} \cdot \text{atm/mole}} \][/tex]
[tex]\[ n = 0.4933990202 \, \text{moles} \][/tex]
Step 4: Round to Appropriate Significant Figures
Given the significant figures in the problem's measurements (2601 mmHg with four significant digits, 3.6 L with two significant digits, and 31°C with two significant digits), the final answer should be rounded to two significant digits.
Therefore, the number of moles of oxygen is:
[tex]\[ n ≈ 0.49 \, \text{moles} \][/tex]
So, the number of moles of oxygen (Oâ‚‚) contained in the cylinder is approximately 0.49 moles.
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
