Answer :
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.
