anonymous
  • anonymous
In an industrial process, the temperature of a fixed amount of gas in a rigid container drops from 50 degrees Celsius to 25 degrees Celsius. What will be the final pressure in the container if the initial pressure is 3.8 atmospheres? 1.9 atm 3.5 atm 4.1 atm 7.6 atm
Chemistry
katieb
  • katieb
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anonymous
  • anonymous
aaronq
  • aaronq
So for this question you would use: \(\sf \large \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\) the temperature needs to be in kelvin though, so add 273 to the temps in celsius.
anonymous
  • anonymous
I have no idea what you mean @aaronq

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aaronq
  • aaronq
Do you know what the question is asking?
anonymous
  • anonymous
nope
aaronq
  • aaronq
lol have you read it? do you know anything from this class..
anonymous
  • anonymous
yea lol I just have a couple questions.
aaronq
  • aaronq
So make an effort and show that you're trying. You need to follow the formula I posted earlier. It compares a closed system, at two instances, when the temp is 50 and 25 celsius.
anonymous
  • anonymous
50 is 323.15 and 25 is 298.15
aaronq
  • aaronq
great. we're also told that the initial pressure is 3.8 atm. At what temp ( 25 or 50) i this true?
aaronq
  • aaronq
In other words, what's the initial temperature?
anonymous
  • anonymous
so I divide 50/3.8 or 323.15/3.8
aaronq
  • aaronq
lets use some algebra to rearrange the whole equation first. \(\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\rightarrow \sf \dfrac{3.8~atm}{323.15~K}=\dfrac{P_2}{298.15~K}\) \(\sf P_2=\dfrac{3.8~atm*298.15~\cancel K}{323.15~\cancel K}\)
aaronq
  • aaronq
now just carryout the operations and you should have the answer
anonymous
  • anonymous
got it thank youu!

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