A gas is at 725 Torr, -8 ° C, and 13.2 L. If the volume is changed to 9,566 ml and the pressure to 1.5atm, wha
A gas is at 725 Torr, -8 ° C, and 13.2 L. If the volume is changed to 9,566 ml and the pressure to 1.5atm, what will the final temperature be?
Update:(A) 77.3 K
(B)156 K
(C)277 K
(D) 302 K
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P1 x V1 / T1 = P2 x V2 / T2
so T2 = (P2/P1) x (V2/V1) x T1
P1 = 725 torr/ 760 = 0.954 atm
V1 = 13.2 L
T1 = -8°C = 265.15 K
P2 = 1.5 atm
V2 = 9,566 mL = 9.566 L
T2 = (1.5/0.954) x (9.566/13.2) x 265.15
T2 = 306.8K = 33.6°C
a million....blended gas regulation.'s regulation: P1 x V1 x T2 = P2 x V2 x T1 a million.0atm x 775mL x 298K = P2 x 615mL x 273. P2 = (a million.0atm x 775mL x 298K) / ( 615mL x 273). P2 = 230,950 / 167,895 = a million.38atm new tension. 2...blended gas regulation. P1 x V1 x T2 = P2 x V2 x T1 a million.5atm x 2.5L x T2 = a million.007atm x a million.5L x 292K T2 = (a million.007atm x a million.5L x 292K) / (a million.5atm x 2.5L) T2 = (a million.007atm x a million.5L x 292K) / 3.seventy 5 = 117.6K - 273 = -a hundred and fifty five.4 °C.