Application of boyle's law a 12-liter tank contains helium gas pressurized to 160 atm. part b what size tank would be needed to contain this same amount of helium at atmospheric pressure (1 atm)? express the size in liters to three significant figures.

According to Mendeleyev-Klapeyron’s equation pV = nRT, where p = 160 atm V = 12 R -constant 0.0821 & T = 298 in Kelvin Using given data, we can determine the amount of Helium gas: n = pV/RT = (160â™12)/(0,0821â™298) = 78,48 (mol) For atmospheric pressure (1 atm) and the same amount we can calculate the volume of tank, using previous equation: V = nRT/p = (78,48â™0,0821â™298)/1 = 1920 (liters) V = 1920 liters Thus Answer is 1920 liters

60.0 - there are 3 significant numbers