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?

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!

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer