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Excalidraw Data
Text Elements
P
atm Pa, kPa mm Hg psi
Pascals,
= 14.7 psi
g
w
654mm
760
1 atm
654/760=.861 atm
879mm
101.3kPa
760mm
117kPa
288.6
-39
0.55 mol of gas = n 105.7 kPa = P 27 degrees celcius = T ( 300 K ) unknown : V
V= nRT/P
PV=nRT
V = (.55)(8.314 LkPa/molk)(300)/105.7) V = .558.314300/105.7 V = 12.97
1 4 5 pg 8 1 4 5 ideal gas law practice ^m3eB1ac2
.52 mol of gas = n 1.3 atm = P 11.8L = V unknown: T
PV=nRT 1.311.8=.52.0821*T 15.34 = .0426T 360.1
10.0 L = V 27 degrees = T ( 300 K ) 3.5 atm = P unknown = n
PV=nRT 3.510.0=n.0821300 35=n.0821*300 35=24.63n 1.42=n
1.42 mol
2.84
2.0
2.3 + 1.7 = 4
434 = 215 + 102 + x 434-215-102 117
17.5 = x + 16.1 1.4 = x
force produced when gas particles collide with container walls and run into a surface
1 atm or 760 mmHg
increase
increase
if variables are on the same side of the equal sign they are indirect if they are on the opposite sides of the equal sign, they are direct
decreases
increases
goes up
increases
increases
n1
n2
P1V1/T1n1 = P2V2/T2n2
P1V1/T1n1 = P2V2/T2n2
these are constant
1.05*2.5 = .980 * x 2.625 = .980 x 2.678 = x
2.7L
V1/T1=V2/T2 10.5/298=x/323 0.035 = x/323 11.38 = x
p1/t1 = p2/t2 .329/320 = x / 350 0.001028125 = x / 350 .36 = x
n1 = .15mol V1 = 2.5L n2 = .55mo V2 = x
2.5/.15 = .55 / x 16.66 = .55 / x 9.1666 = x
V1 = 15.5L P1 = 755mm T1 = 298K unknown = n1
V2 = 22.4L P2 = 760mmHg T2 = 273K unknown = n2
((755)(15.5))/298 = ((760)(x))/273 39.27 = (760x)/273 10720=760x 14.1=x
not always true
2 identical gases at different temperatures same average kinetic energy ? ( no ) same speed ? ( no )
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