Divers know that the pressure exerted by
the water increases about 100 kPa with every
10.2 m of depth. This means that at 10.2 m
below the surface, the pressure is 201 kPa;
at 20.4 m below the surface, the pressure is
301 kPa; and so forth. If the volume of a
balloon is 3.9 L at STP and the temperature
of the water remains the same, what is the
volume 35.78 m below the water’s surface?
Answer in units of L
Suppose that 3.2 L of ethene gas (C2 H4 ) at
1 atm and 298 K is mixed with 8.6 L of oxy-
gen gas at the same pressure and temperature
and burned to form carbon dioxide gas and
liquid water. Ignoring the volume of water
forward, find the final volume of the reaction
mixture (including products and excess reac-
tant) at 1 atm and 298 K if the reaction goes
to completion.
Answer in units of L
For #1, P1V1 = P2V2
For #2,
C2H4 + 3O2 ==> 2CO2 +2H2O
3.2L...8.6L....0.....0
3.2L C2H4 will form 2*3.2 = 6.4L CO2 if it has all of the O2 needed.
8.6L O2 will form 8.6 x (2 mols CO2/3 mols O2) = 5.73 mols CO2 if it has all of the C2H4 needed.
Therefore, O2 is the limiting reagent and you will have 5.73 L O2 formed. We ignore the volume of H2O.l
How much C2H4 is used? That is 8.6L O2 x (1 mol C2H4/3 mol O2) = 2.87 mols used. We had 3.2 initiall; there must be 3.2-2.87 = 0.33L left over.
Total volume = 0.33L C2H4 + 0 oxygen + 5.73L CO2 + ignored H2O = ?
thank you very much! i'll look into redoing the question and seeing if we get the same answer. Thank you for your time
for the first question, how do you find P2. because i understand that the kpa increases by 100 every 10.2 meters but i don't know how to fimd it for 35.78 meters ???
To solve this problem, we need to understand how the pressure affects the volume of the balloon. According to the gas laws, as the pressure increases, the volume decreases (assuming the temperature remains constant).
Given that the pressure increases by 100 kPa with every 10.2 m of depth, we can find the pressure at 35.78 m below the water's surface.
To do this, we can use proportionality. Let's set up a proportion:
Change in pressure / Change in depth = Pressure / Depth
(100 kPa) / (10.2 m) = (Pressure at 35.78 m) / (35.78 m)
Now, we can solve for the pressure at 35.78 m:
(100 kPa) / (10.2 m) = (Pressure at 35.78 m) / (35.78 m)
Cross-multiplying, we get:
(100 kPa) * (35.78 m) = (10.2 m) * (Pressure at 35.78 m)
3580 kPa·m = (10.2 m) * (Pressure at 35.78 m)
Dividing both sides by 10.2 m, we get:
Pressure at 35.78 m = (3580 kPa·m) / (10.2 m)
Pressure at 35.78 m = 351.96 kPa
Now that we have the pressure at 35.78 m, we can calculate the volume of the balloon using Boyle's Law, which states that the pressure and volume of a gas are inversely proportional.
P1 * V1 = P2 * V2
Where:
P1 = Initial pressure (STP pressure = 101.3 kPa)
V1 = Initial volume (3.9 L)
P2 = Final pressure (351.96 kPa, obtained above)
V2 = Final volume (what we need to find)
Plugging in the values, we have:
(101.3 kPa) * (3.9 L) = (351.96 kPa) * (V2)
Now, let's solve for V2:
(101.3 kPa) * (3.9 L) = (351.96 kPa) * (V2)
Cross-multiplying:
(101.3 kPa) * (3.9 L) = (351.96 kPa) * (V2)
393.87 kPa·L = (351.96 kPa) * (V2)
Dividing both sides by 351.96 kPa, we get:
V2 = (393.87 kPa·L) / (351.96 kPa)
V2 ≈ 1.119 L
Therefore, the volume of the balloon at 35.78 m below the water's surface is approximately 1.119 L.