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;
[tex]PV_1 = nRT_1\\\\V_1 = \frac{nRT_1}{P} \\\\[/tex]
where;
R is ideal gas constant, = 0.08205 L.atm./mol.K
[tex]V_1 = \frac{2.5 \times 0.08205 \times 4.2}{1 } \\\\V_1 = 0.862 \ L[/tex]
(a) The volume of the liquid helium at 25 K.
Apply Charles law;
[tex]\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[/tex]
(b) The volume of the liquid helium at 293 K.
[tex]\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[/tex]
