An airtight box, having a lid of area  $80 c{m}^2$ , is partially evacuated ( has lower pressure than outside atmosphere). Atmospheric pressure is $1.01\times {10}^5$ Pa. A force of 600N is required to pull the lid of the box. The pressure in the box is:

Data provided: Input area (A) = 80cm²
Atmospheric pressure = 1.01 × 10⁵ Pa
Power consumption (F active) = 600 N
When F is strong, P presses and A is the valve area
. F_atm = P_atmA
when P_atm is a pressure in space, A is the space cover
as well as
F_inside = P_insideA
When P_inside has pressure in the box, A is the cover area
This replaces the equation F_atm = F_applied + F_inside
Hence  P_atmA = F_applied + P_insideA
$1.01\times {10}^5\times 80\times {10}^{-4}=600+P_{\text{inside }}80\times {10}^{-4}$
$1.01\times {10}^5\times 80\times {10}^{-4}-600=P_{\text{inside }}80\times {10}^{-4}$

 (substituting for $P_{\text{inside }}$)
$P_{\text{inside }}=\frac{\left(1.01\times {10}^5\times 80\times {10}^{-4}-600\right)}{80\times {10}^{-4}}$ (first multiplication and subtraction)
$=1.01\times 80\times 10-60080\times {10}^{-4}=208-60080\times {10}^{-4}=2.6\times {10}^{4}Pa$
Hence the pressure inside the box is $2.6\times {10}^4 \mathrm{Pa}$