The pressure of 9.0 L of an ideal gas in a flexible container is decreased to one-eighth of its original pressure, and its absolute temperature is decreased to one-ninth of the original. What is the final volume of the gas

8L will be the new volume

Explanation:

Let's use the Ideal Gases Law to determine the answer of this question:

Pressure . volume = number of moles . R . T

P . V = n . R . T

We can propose the two situation for the gas:

P₁ . V₁ = n₁ . R . T₁

P₂ . V₂ = n₁ . R . T₂

Notice that R is a constant and n₁ (moles of gas), is not modified with the changes, so we can cancel them.

For the second situation: P₂ = P₁/8 and T₂ = T₁/9

(P₁ . 9L) / T₁ = (P₁/8 . V₂) / T₁/9

(P₁ . 9L) / T₁ . T₁/9 = (P₁/8 . V₂)

P₁ . 1L = P₁/8 . V₂

P₁ . 1L . 8/P₁ = V₂ → 8L

You can also see it, if you put numbers, for example

1 mol of the gas at 1 atm, let's find out the temperature:

1 atm . 9L = 1 mol . 0.082 . T

9/ 0.082 = 110 K

Second situation: 1/8 atm . V = 1 mol . 0.082 . 110/9 K

V = (1 mol . 0.082 . 110/9 K) . 8 atm → 8.02 L ≅ 8L

c. electrical charges of ionic bonds.