Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
To determine the pressure of a gas using the given information, we'll need to follow several steps involving concepts from chemistry, particularly the Ideal Gas Law. The Ideal Gas Law is expressed as:
[tex]\[ PV = nRT \][/tex]
where:
- \( P \) is the pressure in atmospheres (atm),
- \( V \) is the volume in liters (L),
- \( n \) is the number of moles of the gas,
- \( R \) is the ideal gas constant (0.0821 atm·L/(mol·K)),
- \( T \) is the temperature in Kelvin (K).
### Step 1: Convert the given mass of \( \text{N}_2 \) to moles
We start with the mass of nitrogen gas (\( \text{N}_2 \)) which is 40.7 grams. The molar mass of nitrogen gas is calculated as follows:
- Nitrogen (\( \text{N} \)) has an atomic mass of approximately 14.01 g/mol.
- Since \( \text{N}_2 \) consists of two nitrogen atoms, its molar mass is \( 2 \times 14.01 = 28.02 \) g/mol.
Now, we can convert the mass of \( \text{N}_2 \) into moles:
[tex]\[ \text{Moles of } \text{N}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{40.7 \, \text{g}}{28.02 \, \text{g/mol}} \approx 1.452533904 \, \text{moles} \][/tex]
### Step 2: Convert the temperature from Celsius to Kelvin
Next, we need to convert the given temperature from Celsius (\( ^\circ\text{C} \)) to Kelvin (\( K \)):
[tex]\[ T(\text{K}) = T(\text{C}) + 273.15 \][/tex]
[tex]\[ T(\text{K}) = 18.3 + 273.15 = 291.45 \, \text{K} \][/tex]
### Step 3: Use the Ideal Gas Law to calculate the pressure
Now, we can use the Ideal Gas Law to solve for the pressure \( P \):
[tex]\[ P = \frac{nRT}{V} \][/tex]
where:
- \( n = 1.452533904 \) moles,
- \( R = 0.0821 \) atm·L/(mol·K),
- \( T = 291.45 \) K,
- \( V = 5.58 \) L.
Substituting these values in:
[tex]\[ P = \frac{(1.452533904 \, \text{moles}) \times (0.0821 \, \text{atm·L/(mol·K)}) \times (291.45 \, \text{K})}{5.58 \, \text{L}} \][/tex]
[tex]\[ P \approx \frac{(\approx 34.93112354 \, \text{atm·L})}{5.58 \, \text{L}} \][/tex]
[tex]\[ P \approx 6.228726994 \, \text{atm} \][/tex]
### Step 4: Round to the appropriate number of significant figures
The given values (40.7 g, 18.3 \( ^\circ\text{C} \), and 5.58 L) generally have 3 significant figures. Therefore, we will round our final answer for pressure to 3 significant figures:
[tex]\[ P \approx 6.23 \, \text{atm} \][/tex]
### Final Answer
The pressure of 40.7 grams of [tex]\( \text{N}_2 \)[/tex] at [tex]\( 18.3 ^\circ\text{C} \)[/tex] in a 5.58-liter container is [tex]\( 6.23 \)[/tex] atm.
[tex]\[ PV = nRT \][/tex]
where:
- \( P \) is the pressure in atmospheres (atm),
- \( V \) is the volume in liters (L),
- \( n \) is the number of moles of the gas,
- \( R \) is the ideal gas constant (0.0821 atm·L/(mol·K)),
- \( T \) is the temperature in Kelvin (K).
### Step 1: Convert the given mass of \( \text{N}_2 \) to moles
We start with the mass of nitrogen gas (\( \text{N}_2 \)) which is 40.7 grams. The molar mass of nitrogen gas is calculated as follows:
- Nitrogen (\( \text{N} \)) has an atomic mass of approximately 14.01 g/mol.
- Since \( \text{N}_2 \) consists of two nitrogen atoms, its molar mass is \( 2 \times 14.01 = 28.02 \) g/mol.
Now, we can convert the mass of \( \text{N}_2 \) into moles:
[tex]\[ \text{Moles of } \text{N}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{40.7 \, \text{g}}{28.02 \, \text{g/mol}} \approx 1.452533904 \, \text{moles} \][/tex]
### Step 2: Convert the temperature from Celsius to Kelvin
Next, we need to convert the given temperature from Celsius (\( ^\circ\text{C} \)) to Kelvin (\( K \)):
[tex]\[ T(\text{K}) = T(\text{C}) + 273.15 \][/tex]
[tex]\[ T(\text{K}) = 18.3 + 273.15 = 291.45 \, \text{K} \][/tex]
### Step 3: Use the Ideal Gas Law to calculate the pressure
Now, we can use the Ideal Gas Law to solve for the pressure \( P \):
[tex]\[ P = \frac{nRT}{V} \][/tex]
where:
- \( n = 1.452533904 \) moles,
- \( R = 0.0821 \) atm·L/(mol·K),
- \( T = 291.45 \) K,
- \( V = 5.58 \) L.
Substituting these values in:
[tex]\[ P = \frac{(1.452533904 \, \text{moles}) \times (0.0821 \, \text{atm·L/(mol·K)}) \times (291.45 \, \text{K})}{5.58 \, \text{L}} \][/tex]
[tex]\[ P \approx \frac{(\approx 34.93112354 \, \text{atm·L})}{5.58 \, \text{L}} \][/tex]
[tex]\[ P \approx 6.228726994 \, \text{atm} \][/tex]
### Step 4: Round to the appropriate number of significant figures
The given values (40.7 g, 18.3 \( ^\circ\text{C} \), and 5.58 L) generally have 3 significant figures. Therefore, we will round our final answer for pressure to 3 significant figures:
[tex]\[ P \approx 6.23 \, \text{atm} \][/tex]
### Final Answer
The pressure of 40.7 grams of [tex]\( \text{N}_2 \)[/tex] at [tex]\( 18.3 ^\circ\text{C} \)[/tex] in a 5.58-liter container is [tex]\( 6.23 \)[/tex] atm.
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
