At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
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 (O₂) 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 (O₂) contained in the cylinder is approximately 0.49 moles.
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.