Imagine that 10.0 g of liquid helium, initially at 4.20 K, evaporate into an empty balloon that is kept at 1.00 atm pressure. What is the volume of the balloon at (a) 25.0 K and (b) 293 K?

(a) The volume of the liquid helium at 25 K is 5.13 L

(b)  The volume of the liquid helium at 293 K is 60.14 L.

Explanation:

Given;

mass of the liquid helium, m = 10 g

initial temperature of the liquid helium, T₁ = 4.2 K

pressure of the liquid helium, P = 1.00 atm

Atomic mass of Helium, = 4 g

number of moles of Helium, n  = 10 / 4 = 2.5 moles

The initial volume of the liquid helium is calculated as;

where;

R is ideal gas constant,  = 0.08205 L.atm./mol.K

(a) The volume of the liquid helium at 25 K.

Apply Charles law;

(b)  The volume of the liquid helium at 293 K.

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the ima (higher than 1m) will be greater than the ima with the height of 1m.