Question

A 100 cm3 box contains helium at a pressure of 2.20 atm and a temperature of 100 ∘C. It is placed in thermal contact with a 220 cm3 box containing argon at a pressure of 3.70 atm and a temperature of 410 ∘C.

a) What is the initial thermal energy of each gas?

b) What is the final thermal energy of each gas?

c) How much heat energy is transferred, and in which direction?

d) What is the final temperature?

e) What is the final pressure in each box?

Answer 1

a)

in helium box

V = 100 cm^3 = 1*10^-4 m^3

P = 2.2 atm = 2.2*1.013*10^5 pa

T = 100 C = 100 + 273 = 373 k

no of moles of the gas, n_He = P*V/(R*T)

= 2.2*1.013*10^5*1*10^-4/(8.314*373)

= 0.007186 mol

thermal energy = (3/2)*n_He*R*T

= (3/2)*0.007186*373

= 4.02 J <<<<<<<<----------------Answer

b) in Argon box

V = 220 cm^3 = 2.2*10^-4 m^3

P = 3.7 atm = 3.7*1.013*10^5 pa

T = 410 C = 410 + 273 = 683 k

no of moles of the gas, n_Ar = P*V/(R*T)

= 3.7*1.013*10^5*2.2*10^-4/(8.314*683)

= 0.01452 mol

thermal energy = (3/2)*n_Ar*R*T

= (3/2)*0.01452*683

= 14.88 J <<<<<<<<----------------Answer

d) Heat flows from Ar to He untill they attain same temperature.

let T is the final temperature.

(3/2)*n_Ar*R*(683 - T) = (3/2)*n_He*R*(T - 373)

0.01452*(683 - T) = 0.007186*(T - 373)

9.917 - 0.01452*T= 0.007186*T - 2.68

9.917 + 2.68 = T*(0.01452 + 0.007186)

T = (9.917 + 2.68)/(0.01452 + 0.007186)

= 580.3 K or 307.3 C <<<<<<----------Answer

c) heat transfered, dQ = (3/2)*n_Ar*(683 - 580.3)

= (3/2)*0.01452*(683 - 580.3)

= 2.236 J (from Ar to He)

e) at constant volume P/T = constant

for helium box

P2/T2 = P1/T1

P2 = P1*(T2/T1)

= 2.2*(580.3/373)

= 3.422 atm <<<<<<<<----------------Answer

for Argon box

P2/T2 = P1/T1

P2 = P1*(T2/T1)

= 3.7*(580.3/683)

= 3.144 atm <<<<<<<<----------------Answer