2. Two identical rigid boxes contain helium gas. Each box contains 1 mole of hel

Question

2. Two identical rigid boxes contain helium gas. Each box contains 1 mole of helium. We can assume that the helium is an ideal gas, and its heat capacity is Cv 1.5R. Box A is initially at 200°C and Box B is initially at 100°C. The two boxes are brought into thermal contact. The only heat transfer is from Box A to Box B (i.e., the composite system is adiabatic.) (a) (20) How much heat is transferred from Box A to Box B during the process of equilibration and what is the final temperature in the two boxes? (b) (20) What is the net increase in entropy of the universe as the result of this process?

Explanation / Answer

For ideal gas,

Cp - Cv = R

.: Cp = Cv + R = 1.5R + R = 2.5R

(a) By heat balance,

heat lost by box A = heat gained by Box B

nCp(200-T) = nCp(T - 100)

T is the equilibrium temperature.

.: n and Cp is same for both the boxes.

.: 200 - T = T - 100

.: T = 150 C

Heat transfered from box A to box B = nCp(200-150) = 1 mol x 2.5 x 8.314 J/mol-K x 50 C = 1039.25 J

(b) As not heat is transfered to the universe ( since adiabatic)

Enropy change of the universe = 0

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