Using Charles Law, Find T2 when given... read on for more info plz
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Using Charles Law, Find T2 when given... read on for more info plz

[From: ] [author: ] [Date: 11-05-20] [Hit: ]
Thanks a lot!-This is actually a Gay-Lussacs Law problem (constant V).P1/T1 = P2/T2 . . .dont forget that the temperatures must be in Kelvins,......
Hey, i'm having trouble with my Chem HW, i tried this problem a bunch of times and for some reason keep getting it wrong, can you help? Thanks!
The pressure on a gas at -87 degrees C is doubled, but its volume is held constant. What will the final temperature be in degrees C?
Answer in units of Degrees Celsius please. Thanks a lot!

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This is actually a Gay-Lussac's Law problem (constant V).

P1/T1 = P2/T2 . . .don't forget that the temperatures must be in Kelvins, not Celsius.
-87 C + 273 = 186 K

If the pressure is doubled, then P2 = 2P1

P1 / 186 = (2P1) / T2

T2 = (186)(2P1) / (P1) = 372 .. .you double the pressure, the Kelvin temperature also doubles.

372 K - 273 = 99 C

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Are you absolutely certain you need to use Charles's Law? P and T is Gay-Lussac's; a.k.a. pressure/temperature law.

P1/T1=P2/T2 To make it easier to solve for things, rewrite the equation so it's all on one line. (cross multiply)
P1T2=P2T1 Now that you want to solve for T2, divide through by P1 to get:

T2 = (P2T1) / P1

When solving gas law problems, you should convert celcius temperatures to Kelvin (a scale based on absolute 0). The conversion from degree celsius to kelvin is 0C = 273.15 K

T1 = -87 + 273.15 = 186.15 K
Your problem said that the pressure is doubled, so I used 2.00 atm as P1 and 4.00 atm as P2

Plugging into the formula, I get:
T2 = (P2T1) / P1 T2 = (4.00 atm * 186.15K) / 2.00 atm = 372.3 K

Now convert back to celcius by subtracting 273.15 and you should get your answer.
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