at what temp. will 3.6 moles of o2 occupy 55 liters with 2.5 atmospheres pressure?

Answer 1

Ideal Gas Law: PV = nRT

P = pressure in atm = 2.5 atm

V = volume in liters = 55 liters

n = moles = 3.6 moles

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = ?

Solving for T, we have...

T = PV/nR = (2.5 atm)(55 L) / (3.6 mol)(0.0821 Latm/Kmol)

T = 465K

T = 465 - 273 = 192ºC