A manometric fluid dibutyl phthalate has density 1.047x103 kg m-3. what pressure of this liquid of one mm height suggest?
[ans. 0.077 torr]
pls explian!!!
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2 Answers
11P = h (rho) g
= 1 x 10-3 x 1.047 x 103 x 10
= 10.47 N/m2 = 10.47 Pa
Now 133.33 Pa = 1 torr
==> 10.47 Pa = 10.47 / 133.33 = 0.077 torr..
11 torr is the pressure equivalent to 1 mm of Mercury.
from here we can find out the corresponding height of Mercury(equivalent to the pressure exerted by the liquid)
1*1.047*g=h*13.6*g
so h=1.047/13.6=0.0769 mm
1 mm of Hg=1 torr
0.0769 mm of Hg=0.0769 torr
Thus the pressure is 0.769≈0.77 torr.