contestada

What is the final pressure (expressed in atm) of a 3.05 L system initially at 724 mm Hg and 298 K, that is compressed to a final volume of 3.00 L at 273 K?

Respuesta :

Willchemistry

Hey there!:

V1 = 3.05 L

V2 = 3.00 L

P1 = 724 mmHg

P2 = to be calculated

T1 = 298 K

T2 = 273 K

Therefore:

P1*V1  / T1  = P2*V2 / T2

P2 = ( P1*V1 / T1  )   * T2 / V2

P2 = 724 * 3.05 * 273 / 298 * 3.00

P2 = 602838.6 / 894

P2 = 674.31 mmHg

1 atm ----------- 760 mmHg

atm ------------- 674.31 mHg

= 674.31 * 1 / 760

= 0.887 atm

Hope this helps!

Otras preguntas