Answer:
[tex]\large \boxed{\text{103 kPa}}[/tex]
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
[tex]\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}[/tex]
(b) Convert the temperature to kelvins
T = (25 + 273.15) K = 298.15 K
(c) Calculate the pressure
[tex]\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 $\large \boxed{\textbf{103 kPa}}$}[/tex]
