Using the ideal gas law,
<em>PV</em> = <em>nRT</em>
where <em>R</em> = 0.08206 L•atm/(mol•°K), solving for <em>n</em> gives
<em>n</em> = <em>PV</em>/(<em>RT</em>)
<em>n</em> = (845 mmHg) (270 L) / ((0.08206 L•atm/(mol•°K)) (24 °C))
Convert the given temperature to °K and the given pressure to atm:
24 °C = (273.15 + 24) °K ≈ 297.2 °K
(845 mmHg) × (1/760 atm/mmHg) ≈ 1.11 atm
Then the balloon contains
<em>n</em> = (1.11 atm) (270 L) / ((0.08206 L•atm/(mol•°K)) (297.2 °K))
<em>n</em> ≈ 12.3 mol
of He.
Solve the same equation for <em>V</em> :
<em>V</em> = <em>nRT</em>/<em>P</em>
<em />
Convert the target temperature to °K:
-50 °C = (273.15 - 50) °K = 223.15 °K
Then the volume under the new set of conditions is
<em>V</em> = (12.3 mol) (0.08206 L•atm/(mol•°K)) (223.15 °K) / (0.735 atm)
<em>V</em> ≈ 306 L
