h(H2O) = h(Hg) x eHa) = 760 mm(Hg) x 13.5 g/mL = [0,2l00 mm(H20) A barometer with liquid twice as dense as water shows 1 atm as: %3D p(H,0) 1.0 g/mL 0.38 m 1.52 m 5.15 m 20.6 m 1 atm =100 Torr Convert 3.11 x 10-5 atm into mTorr 1 mTorr ? 11 Y 10-5 atm y 16O Torr, 23.0mTorr (AKA: H)
h(H2O) = h(Hg) x eHa) = 760 mm(Hg) x 13.5 g/mL = [0,2l00 mm(H20) A barometer with liquid twice as dense as water shows 1 atm as: %3D p(H,0) 1.0 g/mL 0.38 m 1.52 m 5.15 m 20.6 m 1 atm =100 Torr Convert 3.11 x 10-5 atm into mTorr 1 mTorr ? 11 Y 10-5 atm y 16O Torr, 23.0mTorr (AKA: H)
Chapter7: Statistical Data Treatment And Evaluation
Section: Chapter Questions
Problem 7.16QAP
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Transcribed Image Text:Gas Pressure Conversions
• A given P(atm) takes different column heights (h) of different
liquids (p) to balance in a barometer:
P(H,0) = p(H,O) gn h(H,0)
P(atm) = p(Hg) g, h(Hg)
1
h oc
%3D
%3D
... Find mm(H,0) that measures 1 atm
= 760 mmHg:
%3D
h(H,O) = h(Hg) x e(Ha= 760 mm(Hg) x 13.5_g/mL = [O,2le0
mm(H,O)
%D
%3D
p(H2O)
1.0 g/mL
A barometer with liquid twice as dense as water shows 1 atm as:
0.38 m
1.52 m
5.15 m
20.6 m
Convert 3.11 x 10-5 atm into mTorr
1 atm =100 Torr
3.11 x 10-5 atm xloO Torr
1 atm
1 mTorr
10-3 Torr
23.0mTorr (AKA: µ)
1 in(Hg) = 25.t mm(Hg)
Pa = 29.92 inHg
80.6 KPa
Convert 23.8 inHg into kPa
%3D
1 atm = 760 mmHg = (01325
25.4 mm(Hg)x
1 in(Hg)
%3D
101325 Pa
760 mmHg 1o Pa
23.8 inHg x
x 1 kPa =
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