question_answer 22)

A manometer reads the pressure of a gas in an enclosure as shown in Fig. 7 (NCT) 11 (b) . The liquid used in manometers is mercury and the atmospheric pressure is 76 cm of mercury. (i) Give the absolute and gauge pressure of the gas in the two cases (in units of cm. of mercury). (ii) How would the levels change in case (b) if \[=0\cdot 5\times {{10}^{-3}}{{m}^{3}}.\]cm of water are poured into the right limb of manometer? (Ignore the change in volume of the gas).

Answer:

Here, atmospheric pressure, \[P=76\,cm\] of mercury. (i) In Fig. 7(NCT). 11 (a) pressure head, \[h=+20\,cm\] \[B=\frac{pV}{\vartriangle V}\] Absolute pressure \[=P=76+20=96\,cm\] of mercury Gauge pressure \[h=20\,cm\] of mercury In Fie 7(NCT). 11 (b) pressure head, \[h=-18\,cm\]. Absolute pressure \[=P+h=76+(-18)=58\,cm\] of mercury Gauge pressure \[=h=-18\,cm\] of mercury (ii) Here 13.6 cm of water added in right limb is equivalent to \[\frac{13.6}{13.6}=1\,cm\] of mercury column. i.e. \[=2\cdot 026\times {{10}^{9}}pa\] cm of mercury column. Now pressure at A, \[=1\cdot 0\times {{10}^{5}}pa\] Let the difference in mercury levels in the two lumbs be \[\frac{\text{Bulk modulus of water}}{\text{Bulk modulus of air}}\]then pressure at \[=\frac{2\cdot 026\times {{10}^{9}}}{1\cdot 0\times {{10}^{5}}}\] As, \[=2\cdot 026\times {{10}^{4}}\] \[1\cdot 03\times {{10}^{3}}g{{m}^{-3}}?\] \[=45\cdot 8\times {{10}^{-11}}p{{a}^{-1}}.\] or \[=1\cdot 013\times {{10}^{5}}pa.\]cm of mercury column