Question

955 ml @ 58 celsius and 3.12 atm to 25 celsius and 96 kpa?

Answer 1

@bbcream14

Answer 2

hi@bbcream 14 do u knw charles law?

Answer 3

hmmmm no. I wish i could help but i dont no this. Im sorry:/

Answer 4

its ok thanks anyway

Answer 5

welcome:)

Answer 6

@AccessDenied can u help?

Answer 7

Are we trying to find the new volume?

Answer 8

it says calculate the correct volume

Answer 9

Ah, alright.
This looks to be a case of the combined gas law, though, since we have a pressure temperature and volume all involved.
\( \dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2} \)
There is a lot of units to take care of here.
T_1 and T_2 are supposed to be in Kelvin, so we must convert our temperatures in Celsius into temperatures in Kelvin. All we have to do there is add 273.
Also to note is that the P_1 given is in atm, and the P_2 given is in kPa. We need to convert one or the other to a consistent unit.

Answer 10

can u breakdown further i don't remember this law

Answer 11

It is a combination of Boyle's Law (PV), Charle's Law (V/T) and Guy-Lussac's Law (P/T).
We have information for all but the one variable, V_2, which we want to solve for.
Starting with the left-side (initial gas):
P_1 = 312 atm
V_1 = 955 mL
T_1 = 58 C = 58+273 K Required conversion
----
Now the right side (final gas):
P_2 = 96 kPa *** We must convert this to atm. There are about 101.3 kPa in 1 atm., so
P_2 = 96 / 101.3 = 0.947 atm
V_2 = ? We need to find this.
T_2 = 25 C = 25 + 273 K Required conversion.

Answer 12

So that the information we fill out looks like this:
\( \rm \dfrac{(312 \ atm)(955 \ mL)}{(58+273) \ K} = \dfrac{(0.947 \ atm)V_2}{(25+273) \ K} \)

Answer 13

oh ok i get it thanks srry my server went down

Answer 14

@AccessDenied

Answer 15

If you solve for V_2, the Kelvin units cancel and the atm units cancel. Your remaining unit will be mL. Was I understanding your question properly?

Answer 16

do you multiply 3.12 and 0.947

Answer 17

After seeing the edit, I need to adjust my equation also:
\( \rm \dfrac{(3.12 \ atm)(955 \ mL)}{(58+273) \ K} = \dfrac{(0.947 \ atm)V_2}{(25+273) \ K} \)

We have to solve for V_2 in this case, which means moving the items multiplied onto V_2 to the left-side. To do this, we actually have to apply the inverse operation. For example:
We have to divide both sides by 0.947 atm to get rid of the 0.947 atm multiplied onto V_2.
We have to multiply both sides by (25+273) K to get rid of it in the denominator.

This should end up looking something like this:

Answer 18

So we'll be dividing the 3.12 atm on the left-side by the 0.947 from the right. That way, their units cancel.

Answer 19

i got 2.96

Answer 20

For V_2 ? Or 3.12 / 0.947 ?

Answer 21

for v 2

Answer 22

Hm.. that seems quite low, considering our initial volume was 955 mL.
Starting with:
(3.12 atm)(955 mL) / (58+273 K),
and then we'd be multiplying by the kelvin temperature from the right (25+273 K and dividing by 0.947?
(3.12 atm)(955 mL) / (58+273 K), * (25+273 K) / (0.947 atm)

Answer 23

im confused

Answer 24

What did you do to get 2.96? Can you explain your process and we can see what may have went wrong?

Answer 25

ok so my set up went like this i did (3.12atm)(331)/0.947 atm(298kpa)=3.29/1.11ml

Answer 26

3.12 atm is correctly placed;
331 K is the initial temperature in Kelvin. This should be in the denominaotr / divided off. because PV / T << T is divided.
298 K was the final temperature in Kelvin, but its units seem to be kPa now. It is correctly multiplied on, though. Its units should just be changed.
0.947 atm is correctly placed.
And lastly there is no record of 955 mL in that set up. It should be where 331 is placed.

3.12 atm * 955 mL / 331 K * 298 K / 0.947 atm = V_2 would be correct

Answer 27

so the first part should be 3.12/331?

Answer 28

Yes, although we also need that 955 mL there so that we have a volume unit. All the other units should cancel, but V_2 needs the volume unit from V_1 = 955 mL.

Answer 29

wat?

Answer 30

We were using the combined gas law:
\( \dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{t_2} \)
The set-up involves that V_1 = 955 mL part.
3.12 * \(\color{reD}{955} \) 331/

Answer 31

3.12 * \( \color{red}{955} \) / 331
messed up writing the division of 331

Answer 32

so for the first part its 2,979.6/331=9.00

Answer 33

Yes, that looks good now. :)
Then you multiply by that 298 and divide by 0.947.

Answer 34

2,832.10?

Answer 35

That is around what I got as well. :)
2832 mL

Answer 36

oh ok is that the final anwser

Answer 37

Yes.
V_2 is approx. 2832 mL

Answer 38

ok