In a barometer, air is trapped in the space labelled 'A' above the mercury level in the tube of the barometer, which measures 730 mmHg. If the atmospheric pressure is 750 mmHg, then the pressure of the trapped air is :
Try it first: pick an option, then read the explanation.
Asked in SSC JE 06/06/2024 (09am-11am)
Detailed Explanation
In an ideal Torricellian barometer, the space above the mercury column is a near-perfect Torricellian vacuum (containing only negligible mercury vapor). When air is inadvertently trapped in this space, it exerts a downward pressure that depresses the mercury column.
1. Write the static equilibrium equation at the mercury surface in the cistern:
\[ p_{\text{atm}} = p_{\text{trapped air}} + p_{\text{mercury column}} \]
2. Express all quantities in consistent heads of mercury (\(\text{mmHg}\)):
- Atmospheric pressure \(p_{\text{atm}} = 750\text{ mmHg}\)
- Mercury column height \(h = 730\text{ mmHg}\)
3. Solve for the pressure of the trapped air:
\[ p_{\text{trapped air}} = p_{\text{atm}} - h = 750\text{ mmHg} - 730\text{ mmHg} = 20\text{ mmHg} \]
Why “20 mmHg” Is Correct
The total atmospheric pressure of \(750\text{ mmHg}\) pushing down on the reservoir pool is balanced by the downward weight of the \(730\text{ mmHg}\) liquid column plus the downward force exerted by the trapped gas bubble. Therefore, the trapped gas exerts exactly \(750 - 730 = 20\text{ mmHg}\).
Why the Other Options Are Wrong
- A. 730 mmHg: 730 mmHg is the height of the mercury column itself, not the pressure of the trapped air.
- B. 750 mmHg: 750 mmHg is the ambient atmospheric pressure outside the tube.
- D. 745 mmHg: 745 mmHg has no physical basis and represents an arbitrary subtraction.
Key Point
In a faulty barometer with trapped air: p_atm = p_air + h_Hg ⇒ p_air = p_atm − h_Hg.
Key Concepts for Students
- An ideal Torricellian space has zero gauge pressure and vapor pressure of Hg of only 0.173 Pa at 20°C.
- Trapped air in a barometer causes the instrument to consistently read lower than true atmospheric pressure.
- Head units (mmHg, m of water) represent pressure via p = ρgh and can be subtracted directly when using the same fluid.
Practice more: this is question 3 in Hydrostatic Pressure MCQs – Part 1. Try the whole set and check every answer instantly.