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?

Question

9052 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-01T23:14:43

$P V = n R T$ $PV = nRT$

$\frac{P}{T} = \frac{n R}{V} = \frac{P}{T}$ $P/T = (nR)/V = P/T$

Thus,

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$ $P_1/T_1 = P_2/T_2$

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

Hence,

$\frac{1.60 a t m}{294 K} = \frac{P_{2}}{277 K}$ $(1.60atm)/(294K) = P_2/(277K)$
$therefore P_2 approx 1.51atm$

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