69.5k views
5 votes
Footballs used by the NFL are inflated to 12.5 psi. 11 of 12 footballs measured 10.5 psi after the playoff game when the temperature was 5°C. At what temperature were the balls inflated to explain this pressure drop?

User Flaviu
by
7.9k points

2 Answers

3 votes

Final answer:

After calculating T1, we will have our answer in Kelvin, which can be converted back to degrees Celsius by subtracting 273.15.

Step-by-step explanation:

To determine at what temperature the NFL footballs were inflated before the pressure dropped to 10.5 psi from 12.5 psi, we can use the ideal gas law and assume that the volume of the footballs remains constant (since they're made of materials that don't expand or contract much with temperature). The ideal gas law implies a direct relationship between pressure and temperature, meaning that if the temperature drops, the pressure will also drop, provided the number of moles of gas and the volume remain constant.

Starting with the initial condition, where P1 = 12.5 psi and the final condition where P2 = 10.5 psi at T2 = 5°C, we can find the initial temperature (T1) using the following relation from Charles's Law:

P1/T1 = P2/T2

However, to use this formula, we need to convert the temperatures into Kelvin:


  • T2 (in Kelvin) = 5°C + 273.15 = 278.15 K

  • T1 = (P1 * T2) / P2

T1 = (12.5 psi × 278.15 K) / 10.5 psi

After calculating T1, we will have our answer in Kelvin, which can be converted back to degrees Celsius by subtracting 273.15.

User Raghwendra Sonu
by
7.3k points
1 vote

Answer: 331 K

Step-by-step explanation:

To calculate the temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,


(P_1)/(T_1)=(P_2)/(T_2)

where,


P_1\text{ and }T_1 are the initial pressure and temperature of the gas.


P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:


P_1=12.5psi\\T_1=?\\P_2=10.5psi\\T_2=5^0C=(5+273)K=278K

Putting values in above equation, we get:


(12.5)/(T_1)=(10.5)/(278)\\\\T_1=331K

Hence, the temperature at which the balls inflated to explain this pressure drop is 331 K

User Lynson
by
7.1k points