Answered

Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

If 1.00 mol of argon is placed in a 0.500- L container at 24.0 ∘C , what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation)? For argon, a=1.345(L2⋅atm)/mol2 and b=0.03219L/mol .
Express your answer to two significant figures and include the appropriate units.

Sagot :

pstnonsonjoku

The difference between the ideal pressure and the pressure calculated by the Van Der Waal equation is 2.08 atm.

What is the pressure?

In this problem, we are mandated to obtain the pressure both by the use of the ideal gas equation and then the use of the Van der Walls equation.

Using the idea gas equation;

PV = nRT

P = nRT/V

P = pressure

V = volume

n = number of moles

T = temperature

R = gas constant

P = 1 * 0.082 * (24 + 273)/0.5

P = 48.7 atm

Using the Van Der Wall equation:

P = RT/(V - b) - a /V^2

P = 0.082 * 297/(0.5 - 0.03219) - 1.345/(0.5)^2

P = 24.354/0.46781 - 1.345/ /0.25

P = 52 - 5.38

P = 46.62 atm

The difference between the ideal pressure and the pressure calculated by the Van Der Waal equation is; 48.7 atm - 46.62 atm = 2.08 atm

Learn more about ideal gas equation:https://brainly.com/question/3637553

#SPJ1

We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.