a 6.33-Liter football is filled with air at 1.75 atm at 25.5 degrees Celcius. At the same temperature, the volume of the football is reduced to 2.35 Liters. What is the pressure of air in the ball?
P1V1 = P2V2
To find the pressure of the air in the ball when the volume is reduced, we can use the ideal gas law. The ideal gas law states that the product of pressure, volume, and temperature for a gas is constant. The formula for the ideal gas law is:
PV = nRT
Where:
- P is the pressure of the gas
- V is the volume of the gas
- n is the number of moles of gas
- R is the ideal gas constant (0.0821 L·atm/(mol·K))
- T is the temperature of the gas in Kelvin (K)
In this problem, we are given the initial conditions as:
- Volume (V1) = 6.33 Liters
- Pressure (P1) = 1.75 atm
- Temperature (T1) = 25.5 degrees Celsius
We convert Celsius to Kelvin by adding 273.15:
T1 = 25.5 + 273.15 = 298.65 K
We are also given the final volume (V2) = 2.35 Liters. Since the temperature remains constant (same temperature), we can use the ratio of the initial and final volumes to find the final pressure (P2). The equation can be written as:
(V1 / V2) = (P2 / P1)
Rearranging the equation, we can solve for P2:
P2 = (V1 / V2) * P1
Plugging in the values:
P2 = (6.33 L / 2.35 L) * 1.75 atm
Calculating the value:
P2 = 4.27 atm
Therefore, the pressure of air in the ball when the volume is reduced to 2.35 Liters at the same temperature is approximately 4.27 atm.