A gas at 27°C and 1x103mmHg is cooled in a 3.9 L flask. After its cooled the temp is measured at 11°C. Given that the final pressure was 1.098x103mmHg, what is the new volume of the gas

Answer 1

Given:

T1 = 27ºC + 273 = 300K

P1 =1x10³ mm Hg

V1 = 3.9 L

T2 = 11ºC + 273 = 284K

P2 = 1.098x103 mm Hg

Asked to find:

V2

Using the Combined Gas Law:

P1V1/T1 = P2V2/T2 and substituting values and solving for V2, we have...

(1x10³)(3.9 L)/300 = (1.098x10³)(V2)/284

V2 = 3.4 L