23. The Second Law of Thermodynamics

3 moles of an ideal gas are in the left side of an hourglass-shaped container, separated by a thin barrier. The right side is completely empty, but the volume of the left and right sides are equal. The barrier is suddenly removed, and the gas freely expands into the vacuum. What is the change in entropy?

(III) The specific heat per mole of potassium at low temperatures is given by C_V = aT + bT³, where a = 2.08 mJ/mol · K² and b = 2.57 mJ/mol· K⁴ . Determine (by integration) the entropy change of 0.15 mol of potassium when its temperature is lowered from 5.0 K to 1.0 K.

(III) A general theorem states that the amount of energy that becomes unavailable to do useful work in any process is equal to T_L∆S , where T_L is the lowest temperature available and ∆S is the total change in entropy during the process. Show that this is valid in the specific cases of:

(a) a falling rock that comes to rest when it hits the ground;

(III) A general theorem states that the amount of energy that becomes unavailable to do useful work in any process is equal to T_L∆S , where T_L is the lowest temperature available and ∆S is the total change in entropy during the process. Show that this is valid in the specific cases of:

(b) the free adiabatic expansion of an ideal gas;

(III) Determine the work available in a 2.2-kg block of copper at 490 K if the surroundings are at 290 K. Use results of Problem 57.

At low temperature the specific heat of diamond varies with absolute temperature T according to the Debye equation C_V = 1.88 x 10³ (T/T_D)³ J•mol⁻¹ • K⁻¹ where the Debye temperature for diamond is T_D = 2230 K . Determine the entropy change of 1.00 mol of diamond when it is heated at constant volume from 4 K to 40 K. [Hint: dS = nC _VdT/T , where n is the number of moles.]