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The observed pressure of a 3.029-mole sample of

Kr(g) in a 7.812-L container is 2.867 atm. Use the van der Waals equation to determine the pressure if the gas were behaving ideally.
= 2.32
akr (3)
Ideal pressure =
atm-L²
mol²
atm

The observed pressure of a 3.029-mole sample of Kr(g) in a 7.812-L container is 2.867 atm-example-1
User Noox
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1 Answer

5 votes

Step-by-step explanation:

The van der Waals equation is:

(P + (n^2 a / V^2))(V - n b) = n R T

where P is the observed pressure of the gas, n is the number of moles of the gas, V is the volume of the container, T is the temperature, R is the gas constant, a is the van der Waals constant for the gas, and b is another van der Waals constant for the gas.

To solve for the ideal pressure (P_ideal), we need to use the ideal gas law, which is:

P_ideal = n R T / V

where P_ideal is the pressure that the gas would have if it behaved ideally.

To use this equation, we need to solve for T in the van der Waals equation. Rearranging the equation, we get:

T = (P + (n^2 a / V^2))(V - n b) / (n R)

Substituting the given values, we get:

T = (2.867 atm + ((3^2)(2.32 L^2 atm/mol^2) / (7.812 L)^2))(7.812 L - (3 mol)(0.0391 L/mol)) / (3 mol)(0.0821 L atm/mol K)

T = 88.5 K

Now we can use the ideal gas law to solve for P_ideal:

P_ideal = (3 mol)(0.0821 L atm/mol K)(88.5 K) / 7.812 L

P_ideal = 2.44 atm

Therefore, the pressure that the gas would have if it behaved ideally is 2.44 atm.

User Al Sutton
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