# A basketball is inflated to a pressure of 1.60 atm in a 21.0°C garage. What is the pressure of the basketball outside where the temperature is 4.00°C?

Jan 1, 2018

$P V = n R T$

$\frac{P}{T} = \frac{n R}{V} = \frac{P}{T}$

Thus,

${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$

Note: we must use absolute temperature with all gas law questions.

Hence,

$\frac{1.60 a t m}{294 K} = {P}_{2} / \left(277 K\right)$
$\therefore {P}_{2} \approx 1.51 a t m$

is the pressure of our basketball. No wonder so many tires go flat in Ohio when the temperature drops in the teens!