Question

A very large block of ice, initially at temperature T = 0 oC is placed in a sealed insulated container full of Helium gas, initially at temperature 325 oC and pressure of 3.1 atm. The volume of the Helium is 920 L, and is constant. Helium is a monatomic ideal gas. What is the mass of the liquid water when the system comes to equilibrium? (In other words, how much ice melts?) Assume no heat is lost to the surroundings. Give your answer in kg to at least three significant digits to avoid rounding errors. Do not include units in your answer.

Answer 1

for helium,

T = 325 C = 325 + 273 = 598 k

P = 3.1 atm = 3.1*1.013*10^5 = 3.14*10^5 pa

V = 920 L = 0.92 m^3

use, P*V = n*R*T

n = P*V/(R*T)

= 3.14*10^5*0.92/(8.314*598)

= 58.1 mole

for monatomic gases, Cv = (3/2)*R = (3/2)*8.314 = 12.47 J/(mol K)

now use Heat lost by He as its temperatire decreses from 325 C to 0C,

Q = n*Cv*dT

= 58.1*12.47*325

= 2.355*10^5 J

let m_Ice is the mass of Ice that melts
now use, Q = m_Ice*Lf

==> m_Ice = Q/Lf

= 2.355*10^5/(3.33*10^5)

= 0.707 kg or 707 grams <<<<<<<<<---------------Answer