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Liquid helium at 4.2 K has a density of 0.147 g/mL. Suppose that a 1.00 L metal bottle at 110 K contains air at 1.0 atm pressure. If we introduce 150.0 mL of liquid helium, seal the bottle, and allow the entire system to warm to room temperature (25°C), what is the pressure inside the bottle (in atmospheres)?

User Hrskrs
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1 Answer

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Answer:

The total pressure in the bottle is 119.15 atm

Step-by-step explanation:

Calculate mass of Helium

150ml He * 0.147g/ml = 22.05g He

Calculate moles of Helium

22.05g He / 4g/mole = 5.51 moles

moles air = pv/rt

moles air = 1 atm x 1L / (0.0821L^-atm/mole^-K * 110K) = 0.111 moles air present

if we warm the air alone, p1/t1 = p2/t2

1atm / 110K = xatm / 298K

2.71 atm air pressure

injecting 5.51 moles Helium into 1L flask at 110K

p = nrt/v = 5.51 moles * 0.0821L^-atm/mole^-K * 95K / 1L =42.98atm

pHe: p1/t1 = p2/t2

42.98 atm / 110K = p2 / 298K p2 = 116.44atm

total pressure at 298K = 2.71 + 116.44 atm = 119.15 atm

or

the total pressure at 110K of air and He = 43.98 atm (42.98atm from above for He)

p1/t1 = p2/t2

43.98atm / 110K = xatm / 298K

total pressure = 119.15 atm

The total pressure in the bottle is 119.15 atm

User Wonbyte
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