Chemistry

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I just have a quick question for you.
Do I use Charles' Law to solve the following problem, if not, how should I approach this?

If a 5.0 sample of a gas undergoes an increase in temperature under constant pressure,what is a possible new volume?

Thanks for your help.

  • Chemistry -

    Yes, Charles' Law, assuming 5.0 whatever it is is a volume.

  • Chemistry -

    This gas has undergone 'an increase in temperature under constant pressure'; yet a value has not beeen provided. Please define this for me.

    Thanks again

  • Chemistry -

    What is the 5.0? If that isn't a value, what is it? and what are the units? Or did you make up that value of 5.0 and did not add units to it?

  • Chemistry -

    I'm sorry I don't have an answer for your question..as to what units are involved. In trying to solve,I asked myself the same question. I wrote the question verbatum.

  • Chemistry -

    Incidentally the answer to that chemistry question is suppose to be 5.4L, if that's of any help

  • Chemistry -

    The answer helps a little but not much. The question is a screw-ball one; for it gives no units for the 5.0 and it gives no initial and no final T, just the fact that T increases. I guess the only thing to do is to pick a number (any number?) that is above 5.0 L. An increase in T will mean an increase in volume; therefore, any number above 5.0 should work. But the ONLY way I know to come up with 5.4 L is to assume 5.0 is in L and T increases by by a factor of 0.08. That means IF we start with 298 K, the new T would be 321.84 K. IF we start with 400 K, the new T would be 432 K. I'm sorry I can't be more help. I think this problem has serious problems of its own.

  • Chemistry -

    "I this problem has serious problems of it own" lol
    That comment was great..made me laugh.

    The question was taken from my chemistry text's supplementary online students resource.

    I did email that question to my teacher afterwardds..she taught it was a strange question.

    So, thanks..I appreciate your work.

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