One mole of an ideal gas expands reversibly and isothermally at temperature T until its volume is doubled. The change of entropy of this gas for this process is:
A) R ln 2
B) (ln 2)/T
C) zero
D) RT ln 2
E) 2R

Question

Quizplus · Accepted Answer

Since the process is isothermal, we have 
ΔS = Q/T
where
ΔS = change in entropy
Q = heat absorbed by the gas

Since the gas expands reversibly and isothermally, it absorbs heat from the surroundings. The amount of heat absorbed is given by 
Q = nRT ln(Vf/Vi)
where
n = number of moles
R = gas constant
Vf and Vi are the final and initial volumes, respectively.

In this case, Vi is the volume of 1 mole of gas, and Vf is the volume of 2 moles of gas. Therefore, 
Q = nRT ln(2)

Substituting Q and T in the expression for ΔS, we get 
ΔS = nR ln(2)

Since we have 1 mole of gas, the final answer is 
ΔS = R ln(2)

One Mole of an Ideal Gas Expands Reversibly and Isothermally