Question

The volume of an automobile air bag was 66.8 L when inflated at 25 °C with 77.8 g of nitrogen gas. What was the pressure in the bag in kPa

Answer 1

Answer:

$\large \boxed{\text{103 kPa}}$

Explanation:

We can use the Ideal Gas Law — pV = nRT

Data:

V = 66.8 L

m = 77.8 g

T = 25 °C

Calculations:

(a) Moles of N₂

$\text{Moles of N}_{2} = \text{77.8 g N}_{2} \times \dfrac{\text{1 mol N}_{2}}{\text{28.01 g N}_{2}} = \text{2.778 mol N}_{2}$

(b) Convert the temperature to kelvins

T = (25 + 273.15) K = 298.15 K

(c) Calculate the pressure

$\begin{array}{rcl}pV & =& nRT\\p \times \text{66.8 L} & = & \text{2.778 mol} \times \text{8.314 kPa \cdot L \cdot K ^{-1} mol ^{-1}\times 298.15 K}\\66.8p & = & \text{6886 kPa}\\p & = & \textbf{103 kPa}\end{array}\\\text{The pressure in the bag is \large \boxed{\textbf{103 kPa}} }$