Consider an ideal gas enclosed in a 1.00 L container at an internal pressure of 18.0 atm. If the gas expands against a constant external pressure of 1.00 atm to a final volume of 10.0 L, what is the work done?
Now calculate the work done if this process is carried out in two steps.
1. First, let the gas expand against a constant external pressure of 5.00 atm to a volume of 2.00 L.
2. From there, let the gas expand to 10.0 L against a constant external pressure of 1.00 atm.

Answers

Answer 1

Answer:

A=-0.9117kJ.

B= -0.5065kJ

C=-0.8104KJ

Explanation: A s gas expands, it produces works . Therefore

Work done by gas expansion,

W  = P Δ V  

where P =external  pressure,

ΔV = change in volume (final - Initial)

W = -1 atm (10 - 1 ) L

W =  -9L. atm  x  101.3J/  1L. atm

W = 911.7Joules ≈ -0.9117kJ.

1.The work done in the first step is:

W= - 5.00 atm × (2.00 L - 1.00 L) = -5 atm·L ×(101.3J / 1 atm·L)

=506.5J = -0.5065kJ

2.The work done in the second step is:

w = - 1.00 atm × (10.0 L - 2.00 L) = -8atm·L ×(101.3J / 1 atm·L

=810.4J

w=-0.8104KJ

The work done in the whole  process is:

w =-0.5065kJ --0.8104KJ  = -1.3169KJ


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Please help me, I can't figure this out. You have a 450.0 mL balloon filled with 0.0184 moles of Helium gas. The volume of the balloon will change, but the pressure of the gas inside will always match the 1.00 atm pressure outside. If the balloon is kept at a constant 25oC, how many moles of gas must you remove to shrink the balloon down to 200.0 mL?

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According to Avogadro's law, 0.008 moles of gas must be removed to shrink balloon to 200 ml.

What is Avogadro's law?

Avogadro's law is a gas law which states that the total number of atoms or molecules has a direct proportion to the volume occupied by a gas at a constant temperature and pressure.

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The straight line graph of volume versus moles is a straight line passing through the origin which implies that zero moles of gas occupy zero volume.

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2KCIO3 -2KCL + 302
Use the balanced chemical equation to solve the following problems. Show all work and answer with the correct units and significant figures.
a. How many moles of oxygen are produced if 4 moles of potassium chlorate decomposes?
b. If 4.5 moles of potassium chloride are produced, how many molecules of oxygen gas are also produced?
c. If 5.0 g of potassium chlorate decomposes, how many grams of potassium chloride are produced?

Answers

Answer:

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b. 4.06×10²⁴ molecules of O₂

c. 3.04 g of KCl

Explanation:

Reaction of decomposition is:

2KClO₃  →  2KCl  +  3O₂

a. See stoichiometry value

2 moles of potassium chlorate can decompose to 2 moles of potassium chloride and 3 moles of oxygen. Ratio is 2:3

If 2 moles of KClO₃ can decompose to 3 moles of O₂

Then 4 moles, may decompose to (4 . 3)/2 = 6 moles of O₂

b. In this case, the stoichiometry is the same.

Per 2 moles of KClO₃, I produce 2 moles of KCl

Then, 4.5 moles of KCl, were produced by 4.5 moles of KClO₃

We apply, the last relation:

(4.5 . 3) /2 = 6.75 moles of O₂ are also produced.

How many molecules are in 6.75 moles?

6.75 mol . 6.02×10²³ molecules/mol = 4.06×10²⁴ molecules of O₂

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When the solution is cooled, more [tex]Co(H_2O)_6^{2+}[/tex] (aq) is formed and the equilibrium position shifts towards the left and the solution becomes pink in color.

Balanced equation for the equilibrium:

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