an average airbag contains 132 g of NaN3. How many liters of nitrogen would produce on a warm day at 85 degrees F and a pressure of 1atm?
Follow the steps in this example to find moles N2, then use PV = nRT to solve for volume at the conditions cited.
To determine the volume of nitrogen gas produced, we need to use the Ideal Gas Law equation:
PV = nRT
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 gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature of the gas (in Kelvin, K)
First, let's convert the given temperature from Fahrenheit to Kelvin:
T(K) = (T(°F) - 32) * 5/9 + 273.15
T(K) = (85 - 32) * 5/9 + 273.15
T(K) = 29.444 °C + 273.15
T(K) = 302.594 K
Next, let's calculate the number of moles of NaN3 using the molar mass of NaN3:
Molar mass of NaN3 = (22.99 g/mol + 14.01 g/mol * 3)
Molar mass of NaN3 = 65.00 g/mol
Number of moles (n) = mass / molar mass
Number of moles (n) = 132 g / 65.00 g/mol
Number of moles (n) ≈ 2.03 mol
Now, let's substitute the values into the Ideal Gas Law equation:
PV = nRT
V = (nRT) / P
V = (2.03 mol * 0.0821 L·atm/(mol·K) * 302.594 K) / 1 atm
Calculating the volume:
V ≈ 50.88 L
Therefore, on a warm day at 85 degrees F and a pressure of 1 atm, an average airbag containing 132 g of NaN3 would produce approximately 50.88 liters of nitrogen gas.