If a gas occupies 3.06 liters at 760 mm Hg, what will be the volume at a pressure of 1900 mm Hg?
1 answer:
1.22 litres will be the volume at a pressure of 1900 mm Hg if a gas occupies 3.06 liters at 760 mm Hg.
Explanation:
Data given:
Initial volume of the gas V1 = 3.06 Litres
Initial pressure of the gas P1 = 760 mm Hg
final pressure of the gas P2 = 1900 m Hg
final volume of the gas V2 =?
From the data we can see that Boyle's Law will be applied here,
P1V1 = P2V2
Rearranging the equation, we get
V2 =
putting the values in the equation, we get
V2 =
V2 = 1.224 Litres
The volume is 1.22 litres when the pressure is increased to 1900 mmHg.
You might be interested in
i. The dissolution of PbSO₄ in water entails its ionizing into its constituent ions:
---
ii. Given the dissolution of some substance
,
the Ksp, or the solubility product constant, of the preceding equation takes the general form
.
The concentrations of pure solids (like substance A) and liquids are excluded from the equilibrium expression.
So, given our dissociation equation in question i., our Ksp expression would be written as:
.
---
iii. Presumably, what we're being asked for here is the <em>molar </em>solubility of PbSO4 (at the standard 25 °C, as Ksp is temperature dependent). We have all the information needed to calculate the molar solubility. Since the Ksp tells us the ratio of equilibrium concentrations of PbSO4 in solution, we can consider either [Pb2+] or [SO4^2-] as equivalent to our molar solubility (since the concentration of either ion is the extent to which solid PbSO4 will dissociate or dissolve in water).
We know that Ksp = [Pb2+][SO4^2-], and we are given the value of the Ksp of for PbSO4 as 1.3 × 10⁻⁸. Since the molar ratio between the two ions are the same, we can use an equivalent variable to represent both:
So, the molar solubility of PbSO4 is 1.1 × 10⁻⁴ mol/L. The answer is given to two significant figures since the Ksp is given to two significant figures.