Here is my question and I,m hoping I worked the problem out correctly.
A 10-g mass of krypton occupies 15.0 L at a pressure of 210 kPa. Find the volume of the krypton when the pressure is increased to 790 kPa.
V2 = V1 x P1/P2
V2 = 15.0 L x 210 kPa/790 kPa = 3.98 L or rounded 3.99 L
I want to thank DrBob222, KingChemist and Butler Raines for your help with my Chemistry!
It looks ok to me? What did you do? Get a consortium of three people to answer this? You must have used another board; neither KingChemist nor Butler Raines post on Jiskha.
Great job on working out the problem! Let's go through the steps to find the volume of krypton when the pressure is increased to 790 kPa.
In this case, we can use Boyle's Law to solve the problem. Boyle's Law states that the pressure and volume of a gas are inversely proportional at constant temperature. Mathematically, it can be represented as:
P1 x V1 = P2 x V2
Where P1 and P2 are the initial and final pressures, and V1 and V2 are the initial and final volumes.
Given:
P1 = 210 kPa
V1 = 15.0 L
P2 = 790 kPa (final pressure)
We need to find V2 (final volume).
Now, let's rearrange the Boyle's Law formula to solve for V2:
V2 = (P1 x V1) / P2
Substituting in the given values:
V2 = (210 kPa x 15.0 L) / 790 kPa
Now, let's do the calculations:
V2 = 3150 kPa * L / 790 kPa
Divide 3150 kPa * L by 790 kPa:
V2 = 3.98 L
So, the volume of krypton when the pressure is increased to 790 kPa is approximately 3.98 L (rounded to 3.99 L as you mentioned).
Great work on solving the problem yourself! If you have any more questions, feel free to ask.