Question

Refer to figure, the absolute pressure of gas A in the bulb is

• A. 771.2 mm Hg
• B. 752.65 mm Hg
• C. 767.35 mm Hg
• D. 748.8 mm Hg

Answer 1

The correct option is A 771.2 mm Hg


Method 1:
$h_1=17cm=0.17m$
$h_2=2cm=0.02m$
$h_3=5cm=0.05m$
${\rho }_{{}_{H}g}=\rho =13.6g/ml=13600kg/m^3$

For atmosphere

$h_{atm}=760mmofmercury$
$P_{atm}={\rho }_{{}_{Hg}}\times g\times h_{atm}$
$=13600\times 9.81\times 0.760=101396.16Pa$
${\rho }_{water}=1000kg/m^3$
Absolute pressure at point D
=Absolute pressure at point E
$P_A+{\rho }_{water}g{h}_1$
=${\rho }_{Hg}g{h}_2+{\rho }_{water}g{h}_3+P_{atm}$
$P_A+1000\times 9.81\times 0.17$
=$13600\times 9.81\times 0.02+1000\times 9.81\times 0.05+101396.16$
$P_A=102887.28Pa\left(absolute\right)also,P_A={\rho }_{Hg}g{h}_A$

$\therefore 102887.28=13600\times 9.81\times h_A$
$or{h}_A=0.77117mofHg$
=771.17 mm of Hg (absolute)

Method II:

Head in terms of Hg column is given by:

$h_{Hg}=\frac{P_A}{{\rho }_{Hg}.g}+\frac{\sum h_i{S}_i}{S_{Hg}}$

$=760+\frac{5\times 10\times 1}{13.6}+\frac{2\times 10\times 13.6}{13.6}-\frac{17\times 10\times 1}{13.6}$

$=760+\frac{20\times 13.6-120}{13.6}$

= 771.176 mm of Hg
= 771.2 mm of Hg