Standard Entropy for Liquid at Melting Point

Click For Summary
The discussion focuses on calculating the standard entropy of a liquid at its melting point, determined to be 392.7 K. The vapor pressure equation provided is used to find the pressure of the liquid at this temperature, resulting in a value of 0.8045 mmHg. Using the Clausius-Clapeyron equation, the enthalpy change (ΔH) is calculated as 68361 J/mol, leading to an initial entropy value (ΔS) of 174.07 J/mol K. However, this calculated entropy is noted to be approximately half of the expected value, prompting the consideration of additional variables such as the specific heat capacity (Cp) of 250 J/mol K and the entropy of vaporization at 117 J/mol K. The discussion highlights the need to incorporate these factors for an accurate calculation of standard entropy.
DiffusConfuse
Messages
21
Reaction score
0

Homework Statement


The vapour pressures of a liquid have been measured and fit to the following equation:
Log10 (mmHg) = -3571/T + 6.124
The melting point has been determined to be 392.7 K.
Calculate the standard entropy of the liquid at the melting point.


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
Where's your attempt?
 
My attempt was as follows:
T1= 392.7 K P1 = 0.8045 mmHg (from the equation, giving the vapour pressure of the LIQUID at the melting point).
T2= 298.15 K P2= 1.052E-3 mmHg

From the Clausius Clapeyron:

ln(0.8045/1.052E-3)= ΔH/R*(1/298.15 K - 1/392.7 K)

ΔH/R = 8221.89
ΔH = 68361 J/mol

ΔH/T = ΔS = 68361/392.7 = 174.07 J/mol K

However the answer is about twice this value and a Cp value given for the liquid is 250 J/mol K and the entropy of vaporisation is 117 J/mol K (2 additonal variables)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
525
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
6K