Question

chemistry A basketball is inflated to a pressure of 1.10 atm in a 28.0°C garage. What is the pressure of the basketball outside where the temperature is -2.00°C?

Answer 1

Answer : The final pressure of the basketball is, 0.990 atm

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

`P\propto T`

or,

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

where,

`P_1` = initial pressure = 1.10 atm

`P_2` = final pressure = ?

`T_1` = initial temperature =
`28.0^oC=273+28.0=301.0K`

`T_2` = initial temperature =
`-2.00^oC=273+(-2.00)=271.0K`

Now put all the given values in the above equation, we get:

`(1.10atm)/(P_2)=(301.0K)/(271.0K)`

`P_2=0.990atm`

Thus, the final pressure of the basketball is, 0.990 atm