Question

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?

Answer 1

Answer:

(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;

$PV_1 = nRT_1\\\\V_1 = \frac{nRT_1}{P}$

where;

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

$V_1 = \frac{2.5 \times 0.08205 \times 4.2}{1 } \\\\V_1 = 0.862 \ L$

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

Apply Charles law;

$\frac{V_1}{T_1} =\frac{V_2}{T_2} \\\\V_2 = \frac{V_1T_2}{T_1} \\\\V_2 = \frac{0.862 \times 25 }{4.2} \\\\V_2 = 5.13 \ L$

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

$\frac{V_1}{T_1} =\frac{V_2}{T_2} \\\\V_2 = \frac{V_1T_2}{T_1} \\\\V_2 = \frac{0.862 \times 293 }{4.2} \\\\V_2 = 60.14 \ L$