Question

Calculate the entropy change (ΔS) when one mole of a perfect gas is expanded isothermally and reversibly from an initial pressure 1.00 bar to a final pressure 0.100 bar. Hint: There are two ways to solve this problem. You can follow the in-class exercise did in last week. Also, you can follow the following hint. The temperature of expansion is not given; you cannot know the volume. The pressure keeps changing during the expansion, you have to consider infinitesimal change and use = T , where = p derived from the definition of work and the isothermal process. To solve , you need the perfect gas law and calculate the derivative with respect to p. Once you have the expression of , do integral on both side of the formula (the right hand side is related to the initial and final pressures.)

Answer 1

The entropy change (ΔS) when one mole of a perfect gas is expanded isothermally and reversibly from an initial pressure of 1.00 bar to a final pressure of 0.100 bar is -8.314 J/mol·K * ln(0.100/1.00) = -18.27 J/mol·K.

Step-by-step explanation:

In an isothermal and reversible expansion of a perfect gas, we use the equation for infinitesimal change in entropy dS = (dq_rev)/T. For an ideal gas undergoing isothermal expansion, dq_rev = pdV. Here, dV is an infinitesimal change in volume, and from the ideal gas law, we know that PV = nRT, so V = (nRT)/P. By differentiating this equation with respect to pressure (dV/dP) at constant temperature, we get dV = -(nRT)/(P^2) dP. Substituting this into the expression for dq_rev gives dq_rev = -nRT dP/P. Thus, dS = (-nRT/P) dP/T = -nR dP/P.

Integrating dS from the initial pressure (P_i = 1.00 bar) to the final pressure (P_f = 0.100 bar), we get ΔS = -nR * ln(P_f/P_i) = - (1 mol) * (8.314 J/mol·K) * ln(0.100/1.00) = -18.27 J/mol·K. This negative entropy change indicates the increase in disorder as the gas expands from higher pressure to lower pressure, aligning with the second law of thermodynamics.

The negative sign signifies that the system (the gas) is losing entropy to its surroundings during this process. The calculation involves utilizing the logarithmic relationship between initial and final pressures to determine the entropy change, emphasizing the directional change in entropy due to the expansion of the gas under isothermal conditions.