Question

Two mole of an ideal gas is expanded irreversibly and isothermally at 37°C until its volume is doubled and 3.41 kJ heat is absorbed from surrounding. ∆S total (system + surrounding) is:

A 0
B 0.52J/K
C 22.52J/K
D none of the above

Answer 1

The total change in entropy (∆S total) for the expansion of two moles of an ideal gas that doubles its volume and absorbs 3.41 kJ of heat from the surroundings at 37°C is 0, because the increase in the system's entropy equals the decrease in the surroundings' entropy.

Step-by-step explanation:

The student's question is about the change in entropy (∆S total) when two moles of an ideal gas are expanded isothermally and irreversibly. Given that the expansion doubles the volume and 3.41 kJ of heat is absorbed from the surroundings at a temperature of 37°C (310 K), we use the formula for entropy change ∆S = q_rev/T for the surroundings, where q_rev is the heat absorbed reversibly and T is the absolute temperature. In an irreversible process, the entropy of the surroundings changes by ∆S_surroundings = -q_irr/T, where q_irr is the heat absorbed irreversibly.

Since 3.41 kJ (3410 J) of heat is absorbed, ∆S_surroundings = -3410 J / 310 K = -11 J/K. The entropy of the system increases, because entropy is a state function and is the same for reversible and irreversible processes. So, ∆S_system is positive and equal in magnitude to ∆S_surroundings but with opposite sign (+11 J/K). The total change in entropy is the sum of ∆S_system and ∆S_surroundings, and since they are equal in magnitude but opposite in sign, ∆S_total = ∆S_system + ∆S_surroundings = 0.