Answer :
So, the pressure in the 6.00 L tank with 30.2 grams of nitrogen gas at 385 K is approximately 5.45 atm.
To calculate the pressure of the nitrogen gas in the tank, we can use the ideal gas law equation:
[tex]\[PV = nRT\][/tex]
Where:
- \(P\) is the pressure of the gas (in atmospheres, atm),
- \(V\) is the volume of the gas (in liters, L),
- \(n\) is the number of moles of the gas,
- \(R\) is the ideal gas constant [tex](\(0.0821 \frac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}}\)),[/tex]
- \(T\) is the temperature of the gas (in Kelvin, K).
First, we need to calculate the number of moles of nitrogen gas using the given mass and molar mass of nitrogen:
[tex]\[ n = \frac{m}{M} \][/tex]
Where:
- [tex]\(m\)[/tex] is the mass of the gas (in grams, g),
- [tex]\(M\)[/tex] is the molar mass of the gas (in grams per mole, g/mol).
The molar mass of nitrogen[tex](\(N_2\))[/tex] is approximately 28.02 g/mol.
Given:
- Volume [tex](\(V\))[/tex]= 6.00 L
- Mass [tex](\(m\))[/tex]= 30.2 g
- Temperature [tex](\(T\))[/tex] = 385 K
1. Calculate the number of moles [tex](\(n\)):[/tex]
[tex]\[ n = \frac{30.2 \text{ g}}{28.02 \text{ g/mol}} \][/tex]
[tex]\[ n \approx 1.078 \text{ mol} \][/tex]
2. Plug the values into the ideal gas law equation to find pressure [tex](\(P\)):[/tex]
[tex]\[ P = \frac{nRT}{V} \][/tex]
[tex]\[ P = \frac{(1.078 \text{ mol})(0.0821 \frac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}})(385 \text{ K})}{6.00 \text{ L}} \][/tex]
[tex]\[ P \approx 5.45 \text{ atm} \][/tex]
So, the pressure in the 6.00 L tank with 30.2 grams of nitrogen gas at 385 K is approximately 5.45 atm.
