Final answer:
option a,CH₃OH(l) has a lesser standard molar entropy compared to CH₃OH(g) because the gaseous state allows for more freedom of motion and thus a higher number of available microstates.
Step-by-step explanation:
When comparing CH₃OH(l) and CH₃OH(g), CH₃OH(l) has a standard molar entropy lesser than CH₃OH(g). The key concept here is that entropy, which is a measure of disorder, increases with the degree of freedom of the particles in a substance. In the liquid state, the molecules of CH₃OH are closer together and have less freedom of motion compared to the gaseous state, where the molecules are far apart and move more freely. Therefore, the gaseous state has more microstates available for the molecules, leading to a higher entropy.
When comparing CH3OH(l) and CH3OH(g), we need to consider the phase of the substances. CH3OH(l) refers to liquid methanol, while CH3OH(g) refers to gaseous methanol. In general, gases have higher standard molar entropy compared to liquids because gases have more freedom of movement in terms of molecular motion and are more disordered.
Therefore, CH3OH(g) would have greater standard molar entropy than CH3OH(l).