Arubber ball contains 5.70*10⁻¹ dm³ of gas at a pressure of 1.05 atm. what volume will the gas occupy at 7.47 atm?

Assuming ideal gas behavior:

P1V1 = P2V2

V2 = P1V1/P2 = 1.05 atm * 0.570 liters / 7.47 atm = 0.080 liters

While unproven, it uses facts to support part of a conclusion. it is still factual and has predictive power, however.

Hello! data: p (pressure) = 1.25 atmv (volume) = ? (in l)t (temperature) = 298 kn (number of mols) = 0.900 molesr (gas constant) = 0.082 (atm*l/mol*k)apply the data to the clapeyron equation (ideal gas equation), see: i hope this . =)