a is not right. T must be substituted in Kelvin (K = 273 + C) for Charles' Law to work properly.
2. I worked the problem and obtained 0.0168 L. You are allowed three significant figures so you shouldn't round to two. I think you slipped a decimal point, too.
3. 1.61 atm looks good to me.
Please let me know how I am doing with the following three chemistry questions.
1.A sample of a gas in thermometer has a volume of 135mL at 11 degree celcius.Indicat what temperature(in k and degree celcius)would correspond to the foll.volume:
(a)volume:113mL temperature ??
Solution:Finding final temperature or T2 using Charles'law
T2= V2T1/V1
=(113ML)(11degree celcius/135mL
=9.21degree celcius.
2.If a 5.00L sample of an ideal gas was cooled from 24degree celcius to
-272degree celcius,what will the volume become? Give your answers in sign.figures.
Solution:Applying Charles'law
V2=V1T1/TI
=(5.00)(1kelvin)/297kelvins
= 0.17L
3.A sample of ammonia gas at standard pressure is heated from 20degrees celcius to 200degrees celcius.If the volume remains the same what is the final pressure in mmHg?
Solution:
Finding final pressure or P2,using Gay-Lussac's law
P2=P1T2/T1
=(1atm)(473K)/293K
=1.61atm
Thanks for your work.
1 answer
1 answer