Question

Keeping the number of moles constant, what is the final pressure in atm if the temperature was cooled from -91.5°C to -239°C, the volume decreased from 3.41L to 3.3L and the original pressure was 3990 torr?
Change celcius to Kelvin.

Answer 1

Answer:

The answer to your question is:      P2 = 772.35 torr.

Explanation:

Data

P1 = 3990 torr.

V1 = 3.41 L

T1 = -91.5°C  = 181.5°K

P2 = ?

V2 = 3.3 L

T2 = -239°C = 34°K

Formula

             $\frac{P1V1}{T1} = \frac{P2V2}{T2}$$P2 = \frac{P1V1T2}{T1V2}$

                                 P2 = \frac{(3990)(3.41)(34)}{(181.5)(3.3)}[/tex]

                                 P2 = 462600.6 / 598.95

                                 P2 = 772.35 torr.