Thermodynamics: Conceptual Question

AI Thread Summary
Doubling the number of microscopic states in a thermodynamic system leads to an increase in entropy, as described by the equation S = k*ln(w). The discussion reveals inconsistencies when testing this equation with small versus large numbers of microstates, suggesting that the relationship between microstates and entropy is not linear. Participants emphasize that using small numbers for microstates may not accurately reflect the problem's context, which involves a large number of states. The conclusion is that while entropy increases with more microstates, the exact factor of increase requires careful consideration of the values used. Understanding the logarithmic nature of the relationship is crucial for accurate calculations of entropy changes.
cdlegendary
Messages
15
Reaction score
0
1.Problem: A thermodynamic system has a large number of possible microscopic states. If you double the number of possible microscopic states, what happens to the entropy of the system?

it decreases
it remains the same
it increases, but by a factor less than 2
it increases by a factor greater than 2
it increases by a factor of 2
it increases, but not enough information is given to decide by what factor

2.Equations: S = k*ln(w), where s is entropy, k is the Boltzmann constant, and w is the # of microstates
3.Attempt: I did a couple tests with the equation, subbing in values of microstates. I developed an inconsistency, with it doubling by a factor of two with small numbers, and less than two with larger numbers. So I'm pretty sure the entropy increases, I just don't know by how much. Any help on explaining this to me would be appreciated.
 
Physics news on Phys.org
cdlegendary said:
3.Attempt: I did a couple tests with the equation, subbing in values of microstates. I developed an inconsistency, with it doubling by a factor of two with small numbers, and less than two with larger numbers. So I'm pretty sure the entropy increases, I just don't know by how much. Any help on explaining this to me would be appreciated.
Why bother using small numbers for the microstates? The problem clearly states that the system has a large number of microstates.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top