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

(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.

Step-by-step 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 = (nRT_1)/(P) \\\\`

where;

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

`V_1 = (2.5 * 0.08205 * 4.2)/(1 ) \\\\V_1 = 0.862 \ L`

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

Apply Charles law;

`(V_1)/(T_1) =(V_2)/(T_2) \\\\V_2 = (V_1T_2)/(T_1) \\\\V_2 = (0.862 * 25 )/(4.2) \\\\V_2 = 5.13 \ L`

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

`(V_1)/(T_1) =(V_2)/(T_2) \\\\V_2 = (V_1T_2)/(T_1) \\\\V_2 = (0.862 * 293 )/(4.2) \\\\V_2 = 60.14 \ L`