At the bottom of a container filled with air lies a hollow steel ball with a radius of 2 cm. Ball weight 5g.

At the bottom of a container filled with air lies a hollow steel ball with a radius of 2 cm. Ball weight 5g. To what pressure must the air in the vessel be compressed in order for the ball to rise up? Consider that air at high pressures obeys the equation of the gas state. Air temperature 20 ° C (air compression is rather slow).

Let’s write down the standard value of the molar mass of air:
M = 0.029 kg / mol;
Let’s write down the formulas that we will use when solving:
P * V = (m * R * T) / M;
F ⩾ mg;
F ⩾ pgV;
V = 4/3 * nz ^ 3; 4/3 * nz ^ 3 = (m * R * T) / M;
P = (3 * m * R * T) / (4 * nz ^ 3 * M);
P = (3 * 0.005 * 8.31 * 293) / (4 * 3.14 * 0.02 ^ 3 * 0.029;
P = 12.5 * 10 ^ 6 Pa;
Answer: P = 12.5 * 10 ^ 6 Pa.