One mole of an ideal gas is compressed isothermally at 607.4 K from 5.60 atm to 8.90 atm. If the process is carried out in two irreversible steps (intermediate step at P = 7.00 atm), calculate the following.ΔS

Question

Quizplus · Accepted Answer

The entropy change for an isothermal process is given by:ΔS = -nRln(P2/P1)where n is the number of moles of gas, R is the ideal gas constant, and P1 and P2 are the initial and final pressures, respectively.In this case, we have:n = 1 molR = 8.314 J/mol·KP1 = 5.60 atmP2 = 8.90 atmSubstituting these values into the equation, we get:ΔS = -1 mol * 8.314 J/mol·K * ln(8.90 atm / 5.60 atm) = -3.85 J/KTherefore, the correct answer is B) -3.85 J/K.

One Mole of an Ideal Gas Is Compressed Isothermally at 607.4