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

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

Step-by-step 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

`(P1V1)/(T1) = (P2V2)/(T2)`
`</p><p>                                  P2 = (P1V1T2)/(T1V2)`

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

P2 = 462600.6 / 598.95

P2 = 772.35 torr.