Question

Consider 50 moles of carbon dioxide in a piston-cylinder that is isothermally compressed at temperature of T=350 K from 100 kPa to 400 kPa. What are the changes in the total entropy (of the Universe), the system, and the surroundings if the gas is compressed using a heat reservoir at a temperature of 350 K? Using a heat reservoir at a temperature of 250 K? You can assume that the processes within the piston-cylinder system are internally reversible.

Answer 1

Molecular mass of CO2 is M = 44 g/mol

Mass of 50 moles will be m = 50 * 44 = 2200 g = 2.2 kg

Entropy change of the system = - mR ln (P2 / P1).....where R is gas constant of CO2 = 0.189 kJ/kg/K

= - 2.2 * 0.189 ln (400 / 100)

= - 0.5764 kJ/K

Work done during the process W = - mRT ln (P2 / P1)

= - 2.2 * 0.189 * 350 * ln (400 / 100)

= - 201.74 kJ

Internal energy change () in isothermal process is zero.

By first law of thermodynamics Q - W =

Q - (-201.74) = 0

Q = -201.74 kJ

Heat given to the surroundings is therefore Q_surr = 201.74 kJ

Entropy change of surroundings = Q_surr / T_surr

= 201.74 / 350

= 0.5764 kJ/K

Entropy change of universe = Entropy change of system+surroundings

= -0.5764 + 0.5764

= 0

If the heat reservoir were at 250 K, we will have

Entropy change of surroundings = Q_surr / T_surr

= 201.74 / 250

= 0.807 kJ/K

Entropy change of universe = Entropy change of system+surroundings

= -0.5764 + 0.807

= 0.2306 kJ/K