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.
