November 30, 2022, 06:01:31 PM
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Topic: Gay-Lussac's law  (Read 25776 times)

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Offline Bobbalou

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Gay-Lussac's law
« on: June 30, 2007, 12:18:16 PM »
Hello all,
          I have a few questions to which I thought I had the answer, but they're considered wrong by an answer sheet.

First question is,
                     1) A tightly sealed 5.0-L flask contains 781 mm Hg of Ar at 19 C. The flask is heated until the pressure is doubled. What is the temperature of the gas?

          First, I determined that I will use Gay-Lussac's law in which I will be solving for T2. The equation will be arranged to T2=P2T1/P1. My arithmetic came out
          T2= (1562*19)/781. Units cancel leaving Celcius. My final answer is 38 celcius.
The answer key provided states that the answer is 311C, what gives? I even tried to solve with the constant of volume, but the answer works out to be the same...

Thanks,
         Bobby

Offline iceman

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Re: Gay-Lussac's law
« Reply #1 on: June 30, 2007, 12:35:02 PM »
haha check out the law again, your fractions are upside down. it's P1/T1=P2?T2, try working your problem again, and what does   781 mm Hg .. mean??

Offline Bobbalou

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Re: Gay-Lussac's law
« Reply #2 on: June 30, 2007, 12:51:21 PM »
Thanks,
         I tried it the way you stated but, it's wrong as compared the key. Is there some missing variable?
         As always, here's my work
         P1T1=P2T2 rearranged to (P1T1)/P2=T2
               (781mmHg * 19C)/1562mmHg = T2
               T2 = 9.5C.
         The answer key states it's 311C. That's quite a large answer...

Offline Borek

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Re: Gay-Lussac's law
« Reply #3 on: June 30, 2007, 12:54:48 PM »
Temperature must be in Kelvins, not Celsius.

P1T1=P2T2

Still wrong. See iceman hints.

iceman: mm Hg is just a pressure unit, dated back to times when mercury barometers were used.
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Offline Bobbalou

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Re: Gay-Lussac's law
« Reply #4 on: June 30, 2007, 01:00:15 PM »
LoL, thanks Borek,
      I didn't notice that, it's a typo on the practice test. I'm definitely emailing the professor about this.

*EDIT* Thanks Ice and Borek; as I got 38 C again. I hate when I do this, but I often use a minimal amount of words to convey a point. Btw, the answer key for that problem is
a.   -127 C
b.   38 C
c.   95 C
d.   149 C
e.   311 C (Should be in Kelvin, not C. The key states that this is the answer, but the unit is wrong?)


Offline constant thinker

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Re: Gay-Lussac's law
« Reply #5 on: June 30, 2007, 11:53:41 PM »
Convert to kelvin (add 273), get answer in kelvin, convert answer back to Celsius (subtract 273). Then compare your answer with the answer  key.

In my experience you give the answer in the same unit(s) you were given the initial information in. Sometimes though you are required to convert to other units to solve the problem, just in the end you convert back to the original.
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Offline lsume

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Re: Gay-Lussac's law
« Reply #6 on: September 27, 2015, 05:02:49 AM »
since T must be an absolute value keep the same units so P2=2 X P1 therefore
(2P1/(P1/T1))=T2  2x781/(781/(19+273))=584K=311C

Offline mjc123

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Re: Gay-Lussac's law
« Reply #7 on: September 30, 2015, 08:54:36 AM »
Have 14,000 people really viewed this thread? Were they googling "Gay" for some other reason?

Offline Borek

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Re: Gay-Lussac's law
« Reply #8 on: September 30, 2015, 02:03:54 PM »
Hard to say, queries are not logged (they are part of the HTTP referer, but they are ignored to save on the log space).
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