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,

$\frac{P_1}{P_2}=\frac{T_1}{T_2}$

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:

$\frac{1.10atm}{P_2}=\frac{301.0K}{271.0K}$

$P_2=0.990atm$

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