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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    @aaronq

  2. aaronq
    • one year ago
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    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.

  3. anonymous
    • one year ago
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    I have no idea what you mean @aaronq

  4. aaronq
    • one year ago
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    Do you know what the question is asking?

  5. anonymous
    • one year ago
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    nope

  6. aaronq
    • one year ago
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    lol have you read it? do you know anything from this class..

  7. anonymous
    • one year ago
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    yea lol I just have a couple questions.

  8. aaronq
    • one year ago
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    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.

  9. anonymous
    • one year ago
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    50 is 323.15 and 25 is 298.15

  10. aaronq
    • one year ago
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    great. we're also told that the initial pressure is 3.8 atm. At what temp ( 25 or 50) i this true?

  11. aaronq
    • one year ago
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    In other words, what's the initial temperature?

  12. anonymous
    • one year ago
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    so I divide 50/3.8 or 323.15/3.8

  13. aaronq
    • one year ago
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    lets use some algebra to rearrange the whole equation first. \(\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\rightarrow \sf \dfrac{3.8~atm}{323.15~K}=\dfrac{P_2}{298.15~K}\) \(\sf P_2=\dfrac{3.8~atm*298.15~\cancel K}{323.15~\cancel K}\)

  14. aaronq
    • one year ago
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    now just carryout the operations and you should have the answer

  15. anonymous
    • one year ago
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    got it thank youu!

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