Answer
$$
p_{i}=3.8 \times 10^{4} \mathrm{Pa} \text { . }
$$
Work Step by Step
The magnitude $F$ of the force required to pull the lid off is $F=\left(p_{o}-p_{i}\right) A,$ where $p_{o}$ is the pressure outside the box, $p_{i}$ is the pressure inside, and $A$ is the area of the lid. Recalling that $1 \mathrm{N} / \mathrm{m}^{2}=1 \mathrm{Pa}$ , we obtain
$$
p_{i}=p_{o}-\frac{F}{A}=1.0 \times 10^{5} \mathrm{Pa}-\frac{480 \mathrm{N}}{77 \times 10^{-4} \mathrm{m}^{2}}=3.8 \times 10^{4} \mathrm{Pa} \text { . }
$$
