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

if two electrons have the same n, the same ℓ, and the same ml, then they are in the same orbital. the pauli exclusion principle states that no two identical fermions (a fancy word for electrons and other subatomic particles like electrons) may occupy the same quantum state in an atom simultaneously.