If argon has a volume of 5.0 dm3 and the pressure is .92 atm and the final temperature is 30 degrees celcius and the final volume is 5.7 L and the final pressure is 800 mm Hg, what was the initial temperature of the argon?

Answer 1

(P1V1/T1) = (P2V2/T2)
Remember T must be in kelvin.

Answer 2

5.0 dm^3= 5 Liters

(P1V1/T1)=(P2V2/T2)

P1= .92 atm
V1= 5.0 L
T1=?

P2= (800 mmHg/760 mmHg)= 1.05 atm (There are 760 mmHg in 1 atmosphere)
V2= 5.7 L
T2= 30 degrees Celsius

((.92 atm)*(5.0 L))/(T1)=((1.05 atm)*(5.7 L))/(30 Degrees)

4.6/T1= .1995

4.16/.1995= T1
T1= 20.8 Degrees Celsius

Answer 3

@DrBob222- Temperature does not need to be in Kelvin because it is a ratio, converting to Celsius would give her the same answer.

Answer 4

I beg your pardon. My calculations show two things.
a. your math is not right.
b. your chemistry is way not right.
The ratio of T is NOT the same and T is not the same.
Your method.
(0.92*5/T1) = (1.05*5.7/30)
T1 = (0.92*5*30)/(1.05*5.7)] = 23.06 which I would round to 23.1. I assume degrees C. But this is incorrect.

The correct method.
4.6/T1 = (5.7*1.05/303) = 0.01975
T1 = 0.01975/4.6 = 232.88 K which is 232.88-273.15 = -40.26 C.
BIG DIFFERENCE.
Temperature MUST be in kelvin.

Answer 5

how do i find out what the normal boiling point of a liquid could be if the vapor pressure of the liquid is .92 atm at 60 degrees celsius?

Answer 6

skeet

Answer 7

well okay lol