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: 103 kpa

Step-by-step explanation:

Answer 2

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

Step-by-step 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) * \frac{\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 * \text{66.8 L} & = & \text{2.778 mol} * \text{8.314 kPa$\cdot$ L$\cdot$K$^(-1)$mol$^(-1)*$ 298.15 K}\\66.8p & = & \text{6886 kPa}\\p & = & \textbf{103 kPa}\end{array}\\\text{The pressure in the bag is $\large \boxed{\textbf{103 kPa}}$}`