Final answer:
The pressure to which 7200 liters of air must be compressed to fit into an 80.0 cm wide spherical air tank is approximately 26.866 atmospheres.
Step-by-step explanation:
To calculate the pressure to which the volume of air must be compressed in order to fit into an 80.0 cm wide spherical air tank, we can use the ideal gas law in combination with the volume of the sphere. The formula for the volume of a sphere is V = (4/3)πr^3, where V is volume and r is radius. First, we need to convert the width of the sphere (which is its diameter) from centimeters to meters to find its radius in meters, then to calculate the volume in cubic meters.
Radius: 80.0 cm / 2 = 40.0 cm = 0.40 m
The volume (V) of the tank is then: V = (4/3)π(0.40)^3 = 0.268 m^3
Converting the air needed from liters to cubic meters: 7200 L = 7.2 m^3.
According to the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the amount of substance (moles), R is the ideal gas constant, and T is temperature. In this case, assuming temperature and amount of substance are constant, the pressure can be related to the volume by P1V1 = P2V2 (Boyle's Law). Since we don't need to work out the intermediate steps of the gas constants, we relate the initial and final pressures and volumes directly.
Let V1 = 7.2 m^3 (air needed), and V2 = 0.268 m^3 (volume of the tank).
The initial pressure (P1) is 1 atmosphere since it's the pressure of the air in standard conditions. We need to find the final pressure (P2).
P1V1 = P2V2
P2 = (P1V1) / V2
P2 = (1 atm × 7.2 m^3) / 0.268 m^3
P2 = 26.866 atm (rounded to three significant digits)
The pressure to which the 7200 liters of air must be compressed to fit into the spherical air tank is about 26.866 atmospheres.