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.
[tex]P\propto T[/tex]
or,
[tex]\frac{P_1}{P_2}=\frac{T_1}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure = 1.10 atm
[tex]P_2[/tex] = final pressure = ?
[tex]T_1[/tex] = initial temperature = [tex]28.0^oC=273+28.0=301.0K[/tex]
[tex]T_2[/tex] = initial temperature = [tex]-2.00^oC=273+(-2.00)=271.0K[/tex]
Now put all the given values in the above equation, we get:
[tex]\frac{1.10atm}{P_2}=\frac{301.0K}{271.0K}[/tex]
[tex]P_2=0.990atm[/tex]
Thus, the final pressure of the basketball is, 0.990 atm
