A sample of methane occupies a volume of 370.0 ml at 25 oC and exerts a pressure of 1020 mm Hg. If the volume is allowed to expand to 510.0 ml at temperature of 300 K, What will be the pressure?

744.9 mmHg ≅ 745 mmHg

Explanation:

The base to solve this, is the Ideal Gases Law. The mentioned formula is:

P . V = n . R . T

To compare two situations, we can propose:

For the first situation P₁ . V₁ = n₁. R . T₁

For the second situation P₂ . V₂ = n₂ . R . T₂

As the sample has the same moles and R is a constant value, we can avoid them so: (P₁ . V₁) / T₁ = (P₂ . V₂) / T₂

We need to make Tº unit conversion:

25ºC + 273 = 298K

We replace data → (370 mL . 1020 mmHg) / 298K = (P . 510 mL) / 300 K

(377400 mL.mmHg / 298K) . 300 K = P . 510 mL

379932.8 mL . mmHg = P . 510 mL

(379932.8 mL . mmHg) / 510 mL = P → 744.9 mmHg

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explanation:

metric time was an attempt to create a standard unit of time based off of 10, but it never took off an is not included in the metric system (si)