Question

Liquid helium at 4.2 K has a density of 0.147 g/mL. Suppose that a 2.00-L metal bottle that contains air at 114K and 2.0 atm pressure is sealed off. If we inject 130.0 mL of liquid helium and allow the entire system to warm to room temperature (25°C), what is the pressure inside the bottle?

Answer 1

mass of He = 130 mL *0.147g/mL = 19.11 gm

moles of He = 19.11gm/4gm mol^-1 = 4.78 moles

moles air = PV/RT
moles air = 2atm x 2L / (0.0821L-atm/mole-K * 114K) = 0.42moles air present
if we warm the air alone, P1/T1 = P2/T2
2atm / 114K = xatm / 298K
x = 5.22 atm

injecting 4.78moles He into 2L flask at 114K
p = nrt/v = 4.78moles x 0.0821L-atm/mole-K * 114K / 2L = 22.34atm
pHe: p1/t1 = p2/t2
22.34atm /114K = p2 / 298K, p2 = 58.40atm

the total pressure at 114K of air and He = 2 atm + 22.34 atm = 26.34 atm

total pressure at 298 K = 58.40 +5.22 = 63.62 atm.