During laparoscopic surgery, carbon dioxide gas is used to expand the abdomen to help create a larger working space.
If 5.15 L of CO2 gas at 21 ∘C at 793 mmHg is used, what is the final volume, in liters, of the gas at 33 ∘C and a pressure of 749 mmHg , if the amount of CO2 remains the same?
I have it set up to V1*p2/T1=V2*p2/T2. My problem is the algebra. I am solving for V2. However I don t know how to get rid of a numerator
my problem: (5.15L)(793mmHg)/294K=V2(749mmHg)/306K
Please help thanks!:)
Chemistry question:?
2016-05-27 7:56 pm
回答 (5)
2016-05-28 2:42 am
first of all.. the BETTER way of solving this is..
(1) start with the equation.. P1V1 / (n1T1) = P2V2 / (n2T2)
(2) rearrange for your desired unknown
(3) identify anything held constant and cancel it
(4) plug and chug. remember.. T in K or R.. never °C nor °F
**********
this problem
starting with
.. P1V1 / (n1T1) = P2V2 / (n2T2)
rearranging
.. V2 = V1 x (P1 / P2) x (T2 / T1) x (n2 / n1)... <== note the grouping !
since "amount of CO2 remains the same".. n2 = n1 and n2/n1 = 1... that term drops out!
.. V2 = V1 x (P1 / P2) x (T2 / T1)
plug and chug
.. V2 = 5.15L x (793mmHg / 749mmHg) x ((33+273.15 K) / (21+273.15 K))
.. .. . = 5.67L
********
yes... you can cross multiply after you plug in the data like one of the other answerers recommended... but the method I showed you is MUCH MORE ACCURATE. Manipulate the variables FIRST.. you should be able to eyeball that P1V1 go together.. so P2 must be in the denominator.. that V2 and n2 must be in the numerator. Easy eyeball check of the setup... THEN sort out what V1 P1 P2 T2 T1 are and plug them in.
AND you get the added benefit that this process works for ALL those two phase gas law problems.. you don't have to remember boyles law, Charles law, combined law, etc. They are all derivatives of the ideal gas law with different things held constant. And we're canceling those out anyway.
for more on this
https://answers.yahoo.com/question/index?qid=20140321080145AAohDWo
https://answers.yahoo.com/question/index?qid=20140304081407AAuHZjf
(1) start with the equation.. P1V1 / (n1T1) = P2V2 / (n2T2)
(2) rearrange for your desired unknown
(3) identify anything held constant and cancel it
(4) plug and chug. remember.. T in K or R.. never °C nor °F
**********
this problem
starting with
.. P1V1 / (n1T1) = P2V2 / (n2T2)
rearranging
.. V2 = V1 x (P1 / P2) x (T2 / T1) x (n2 / n1)... <== note the grouping !
since "amount of CO2 remains the same".. n2 = n1 and n2/n1 = 1... that term drops out!
.. V2 = V1 x (P1 / P2) x (T2 / T1)
plug and chug
.. V2 = 5.15L x (793mmHg / 749mmHg) x ((33+273.15 K) / (21+273.15 K))
.. .. . = 5.67L
********
yes... you can cross multiply after you plug in the data like one of the other answerers recommended... but the method I showed you is MUCH MORE ACCURATE. Manipulate the variables FIRST.. you should be able to eyeball that P1V1 go together.. so P2 must be in the denominator.. that V2 and n2 must be in the numerator. Easy eyeball check of the setup... THEN sort out what V1 P1 P2 T2 T1 are and plug them in.
AND you get the added benefit that this process works for ALL those two phase gas law problems.. you don't have to remember boyles law, Charles law, combined law, etc. They are all derivatives of the ideal gas law with different things held constant. And we're canceling those out anyway.
for more on this
https://answers.yahoo.com/question/index?qid=20140321080145AAohDWo
https://answers.yahoo.com/question/index?qid=20140304081407AAuHZjf
2016-05-27 8:15 pm
(5.15 L) × (793 mmHg) / (294 K) = V₂ (749mmHg) / (306 K)
Reverse the two sides :
V₂ (749mmHg) / (306 K) = (5.15 L) × (793 mmHg) / (294 K)
Eliminate the same units on the both sides :
V₂ × 749 / 306 = (5.15 L) × 793 / 294
Multiply the both sides by (306 / 749) :
[V₂ × 749 / 306] × (306 / 749) = [(5.15 L) × 793 / 294] × (306 / 749)
Then :
V₂ = [(5.15 L) × 793 / 294] × (306 / 749)
V₂ = 5.68 L
Reverse the two sides :
V₂ (749mmHg) / (306 K) = (5.15 L) × (793 mmHg) / (294 K)
Eliminate the same units on the both sides :
V₂ × 749 / 306 = (5.15 L) × 793 / 294
Multiply the both sides by (306 / 749) :
[V₂ × 749 / 306] × (306 / 749) = [(5.15 L) × 793 / 294] × (306 / 749)
Then :
V₂ = [(5.15 L) × 793 / 294] × (306 / 749)
V₂ = 5.68 L
2016-05-27 8:49 pm
I see now. I was stumped on the algebra part. Thankyou everyone who is willing to help!
2016-05-27 8:05 pm
The easiest thing to do is to multiply both sides by 306K to give:
5.15 X 793 X 306/294 = V2 (749)
Are you OK from there?
5.15 X 793 X 306/294 = V2 (749)
Are you OK from there?
2016-05-27 8:28 pm
There are several correct methods, but about 1000 incorrect ones as well.
Any time you have a fraction = another fraction ... then you can cross multiply ... all of 1st numerator X all of 2nd denominator = all of 1st denominator X all of 2nd numerator
(5.15L)(793mmHg)/294K=V2(749mmHg)/306K
V2(749mmHg)(294K) = (5.15L)(793mmHg)(306K)
.... notice mmHg and K will cancel leaving just Liters for units of the answer
now divide by all coefficients of the variable (V)
V = [(5.15L)(793mmHg)(306K)] / [(749mmHg)(294K)]
now just use a calculator
Any time you have a fraction = another fraction ... then you can cross multiply ... all of 1st numerator X all of 2nd denominator = all of 1st denominator X all of 2nd numerator
(5.15L)(793mmHg)/294K=V2(749mmHg)/306K
V2(749mmHg)(294K) = (5.15L)(793mmHg)(306K)
.... notice mmHg and K will cancel leaving just Liters for units of the answer
now divide by all coefficients of the variable (V)
V = [(5.15L)(793mmHg)(306K)] / [(749mmHg)(294K)]
now just use a calculator
收錄日期: 2021-04-18 14:53:50
原文連結 [永久失效]:
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