A 610-g quantity of an ideal gas undergoes a reversible isothermal compression at a temperature of 330 K. The compression reduces the volume of the gas from 0.40 m³ initially, to 0.21 m³ finally. The molecular mass of the gas is 33.0 g/mol, and the ideal gas constant is R = 8.314 J/(mol∙K). The entropy change for the gas is closest to A) -99 J/K. B) -81 J/K. C) 99 J/K. D) 81 J/K. E) 0.00 J/K.

Question

Quizplus · Accepted Answer

Since the process is isothermal, the temperature remains constant throughout the compression, and thus we can use the equation for entropy change in an ideal gas undergoing reversible isothermal compression as follows: ΔS = nR ln(V2/V1) Where n is the amount of substance (in moles) of the gas, and V1 and V2 are the initial and final volumes, respectively. First, we need to calculate the amount of substance of the gas using the given mass and molecular mass: n = m/M = 610 g / 33.0 g/mol = 18.48 mol Substituting the given values into the equation for entropy change, we obtain: ΔS = (18.48 mol) (8.314 J/(mol∙K)) ln(0.21 m³/0.40 m³) ≈ -98.7 J/K Therefore, the closest answer choice is A) -99 J/K.

A 610-G Quantity of an Ideal Gas Undergoes a Reversible