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24 April, 21:41

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

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  1. 24 April, 22:50
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    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
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