Manometer

1. Liquid at A is exposed to atmosphere. Thus, pressure $P_A$ at A is the atmospheric pressure $P_{atm}$
2. When gas supply is turned on, gas will exert pressure $P_B$ on liquid at B. As $P_B$ is higher than the atmospheric pressure, liquid at A will rise until pressure $P_C$ at C, which is on the same horizontal level as B, equals the gas pressure. Pressure $P_C$ at C is given by $P_C-P_A=h\rho g$
3. Since $P_C=P_B$ and $P_A=P_{atm}$, we determine pressure $P_B$ of gas to be: $P_B-P_{atm}=h\rho g$ $P_B=h\rho g+P_{atm}$

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For manometer with gas pressure < atmospheric pressure → formula is $P_{gas}+\Delta h\rho g=P_{atm}$ → $P_{gas}=P_{atm}-\Delta h\rho g$

• if both ends are exposed to atmospheric pressure ⇒ then liquids will be at the same level. Calculate the average of both heights, that is the height of both sides now

• Explain why water cannot be used

  • Water has a much lower density than mercury. This means that a very long tube needs to be used in the manometer and this is not practical. +++

Level of liquid in manometer ⇒ is independent of cross sectional area of tube

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