A sample of Ar gas has a volume of 6.30 L with an unknown pressure. The gas has a volume of 8.57 L when the pressure is 2.74 atm, with no change in temperature or amount of gas. What was the initial pressure, in atmospheres, of the gas?

3.73 atm

Explanation:

Since there is no change in temperature, we use Boyle's law.

P₁V₁ = P₂V₂ where P₁ = initial pressure of Ar gas = unknown, P₂ = final pressure of Ar gas = 2.74 atm, V₁ = initial volume of Ar gas = 6.30 L and V₂ = final volume of Ar gas = 8.57 L

P₁ = P₂V₂/V₁ = 2.74 atm × 8.57 L/6.30 L = 3.73 atm

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.