arles\'s Law For a fixed quantity of a gas at constant pressure calculate the fi
ID: 706205 • Letter: A
Question
arles's Law For a fixed quantity of a gas at constant pressure calculate the final/initial volume/temperature if the initial/final temperature/volume were the initial/natnenature the gas would occupy changed as shown in the following ta Final Initial Initial Final Volume Temperature Temperature Volume 6.00L 0.0 °C 15:41? | 140.0 mL 10130 |(070. 84 nul 30.0 °C | 70.0 mL 0.0 ? 212 mL 0.0? 401.0 °C 2.55 L -37 °C 85 mL 20S13C 47.5L 212 K SIL 337K 87.5L 165 K 135LH.SK 100.0 ? 224 L 0.0°C 3% ,OSL/ 100.0 mL -n 250.0 ml. 100.0 °C 450 K 152 mL 45 K 182.74?| 148 mL | 125 mL 225? 27.5°C 391.UB | 450? | 156 mL 125255 13.7L 45 °C S72L 125053 300 KExplanation / Answer
1) from charles law,
V1/T1 = V2/T2
V1 = initial volume = 6.00 L
T1 = initial temperature in kelvins = 30.0+273.15 = 303.15 k
V2 = final volume = ?
T2 = final temperature in kelvins = 0.0+273.15 = 273.15 k
(6/303.15) = (V2/273.15)
v2 = final volume of gas = 5.406 L
answer: 5.41 L
2) from charles law,
V1/T1 = V2/T2
V1 = initial volume = 70.0 ml
T1 = initial temperature in kelvins = 0.0+273.15 = 273.15 k
V2 = final volume = 140.0 ml
T2 = final temperature in kelvins = ? k
(70/273.15) = (140.0/T2)
T2 = temperature in kelvins = 546.3 K
in degree celcius = 546.3-273.15 = 273.15 C
answer: 273.15 C
3)
from charles law,
V1/T1 = V2/T2
V1 = initial volume = 212 ml
T1 = initial temperature in kelvins = -60.0+273.15 = 213.15 k
V2 = final volume = ?
T2 = final temperature in kelvins = 401.0+273.15 = 674.15 k
(212/213.15) = (V2/674.15)
v2 = final volume of gas = 670.51 ml
answer: 670.51 ml
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