What Are the Van der Waals Constants for Sulfur Hexafluoride (SF6)?

AI Thread Summary
The van der Waals constants for Sulfur Hexafluoride (SF6) are a = 6.871 L² bar/mol² and b = 0.0562 L/mol. The critical temperature (Tc) is 318.7 K and the critical pressure (Pc) is 37.6 bar. These constants are essential for understanding the intermolecular forces within SF6 and are used in the van der Waals equation to analyze gas behavior under varying conditions. The information is crucial for drawing critical points and isothermal curves in laboratory settings. Accurate values for these constants facilitate better predictions of gas behavior at high pressures and low temperatures.
diegojco
Messages
23
Reaction score
0
Could anyone here say me what's the constants of van der waals for the Sulfur Hexaflouride (SF6)
 
Physics news on Phys.org
please anyone can help me? it is so urgent since this is for draw the critical point and the critical isothermal for the laboratory. Or you can say me what's the critical molar volume or the critical volume for this gas, because I have the critical pressure and temperature.
 


The van der Waals constants for Sulfur Hexafluoride (SF6) are as follows:

a = 6.871 L^2 bar/mol^2
b = 0.0562 L/mol
Tc = 318.7 K
Pc = 37.6 bar

These constants describe the intermolecular forces between SF6 molecules, and are used in the van der Waals equation to calculate the behavior of gases at high pressures and low temperatures.
 

Thread 'Errors and significant figures'

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...
View full post »
Thread 'A cylinder connected to a hanging mass'

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...
View full post »

Similar threads

Verifying properties of Van der Waals Gas
Replies
2
Views
1K
Question in Thermal Physics (Van der Waals' Equation)
Replies
6
Views
2K
Is Entropy decreased for Free expansion of a Waals gas?
Replies
6
Views
3K
Find thermal capacity of a Van der Waals gas
Replies
31
Views
1K
About Van Der Waals interactions
Replies
2
Views
2K
Calculate the pressure using the Van der Waal equation on reduced variables
Replies
6
Views
1K
Entropy change of van der Waals gas expansion
Replies
10
Views
21K
Finding b in Van der Waals Equation?
Replies
9
Views
4K
A I'm trying to follow the proof given in Box 5.3 of MCP (Thorne/Blandford)
Replies
4
Views
2K
What is initial slopes of the plot Z versus P ?
Replies
3
Views
1K

Hot Threads

Recent Insights

Back
Top