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It is desired to inflate a baggie with a volume of 816 milliliters by filling it with nitrogen gas at a pressure of 0.847 atm and a temperature of 303 K. How many grams of N2, gas are needed? _____grams

User Triforceofcourage
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1 Answer

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Step-by-step explanation:

We have to inflate a baggie and we know the volume, the pressure, and the temperature of the gas. The formula that relates all these measures is the ideal gas law.

P * V = n * R * T

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas contant (0.082 atm*L/mol*K) and T is the temperature.

Before we plug the values in the formula we have to convert the volume from mL to L.

1000 mL = 1 L

V = 816 mL * 1 L/(1000 mL) = 0.816 L

V = 0.816 L

P = 0.847 atm

T = 303 K

We can use the given values and solve the formula for the number of moles.

P * V = n * R * T

n = P * V/(R * T)

n = 0.847 atm * 0.816 L/(0.082 atm*L/(mol*K) * 303 K)

n = 0.0278 moles

Now we can convert those 0.0278 moles of N₂ into grams using the molar mass of N₂.

molar mass of N = 14.00 g/mol

molar mass of N₂ = 2 * 14.00 g/mol

molar mass of N₂ = 28.00 g/mol

mass of N₂ = 0.0278 moles * 28.00 g/mol

mass of N₂ = 0.778 g

Answer: 0.778 g of nitrogen are needed.

User Jucajl
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