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1. A faulty barometer tube has some air at the top above the mercury. When the length of the air column is 250 mm , the reading of the mercury above the outside level is 750 mm . When the length of the air column is decreased to 200 mm , the reading of the mercury above the cutside level becomes 746 mm . Calculate the atmospheric pressure.

Ask by Delgado Howell. in Nigeria
Feb 03,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The atmospheric pressure is 766 mm Hg.

Solution

To calculate the atmospheric pressure using the given data from the faulty barometer, let's analyze the situation step by step. ### Understanding the Problem The barometer consists of a mercury column with trapped air above it. When the length of the trapped air column changes, the height of the mercury column adjusts accordingly. The atmospheric pressure (\( P_{\text{atm}} \)) is balanced by the pressure exerted by the trapped air and the mercury column. ### Given Data 1. **First Scenario:** - Length of air column (\( L_1 \)) = 250 mm - Mercury column height (\( h_1 \)) = 750 mm 2. **Second Scenario:** - Length of air column (\( L_2 \)) = 200 mm - Mercury column height (\( h_2 \)) = 746 mm ### Assumptions - **Temperature remains constant**, so the pressure and volume of the trapped air are inversely related (Boyle’s Law): \( P_1 \times L_1 = P_2 \times L_2 \). - **Pressure is measured in mm Hg**, assuming the mercury column's height directly correlates to pressure. ### Formulating the Equations From the balance of pressures: \[ P_1 + h_1 = P_{\text{atm}} \quad \text{and} \quad P_2 + h_2 = P_{\text{atm}} \] Given Boyle’s Law: \[ P_1 \times L_1 = P_2 \times L_2 \] Substituting the pressures from the balance equations: \[ (P_{\text{atm}} - h_1) \times L_1 = (P_{\text{atm}} - h_2) \times L_2 \] Plugging in the known values: \[ (P_{\text{atm}} - 750) \times 250 = (P_{\text{atm}} - 746) \times 200 \] ### Solving for \( P_{\text{atm}} \) Expand and rearrange the equation: \[ 250P_{\text{atm}} - 187500 = 200P_{\text{atm}} - 149200 \] \[ 250P_{\text{atm}} - 200P_{\text{atm}} = 187500 - 149200 \] \[ 50P_{\text{atm}} = 38300 \] \[ P_{\text{atm}} = \frac{38300}{50} = 766 \text{ mm Hg} \] ### Conclusion The atmospheric pressure is **766 mm Hg**. **Answer:** The atmospheric pressure is 766 millimeters of mercury

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The Deep Dive

To calculate the atmospheric pressure, we can use the barometric formula, which states: P_atm = P_0 + ρgh Where: - P_atm is the atmospheric pressure, - P_0 is the pressure exerted by the air column, - ρ is the density of mercury (approximately 13,600 kg/m³), - g is the acceleration due to gravity (approximately 9.81 m/s²), - h is the height of the mercury column. From the two scenarios given in the problem, we can calculate the effective pressure exerted by the air column at different lengths. In the first case with a 250 mm air column, the height of mercury is 750 mm: Using the formula: P_0 (250 mm) = 750 mm + 250 mm = 1000 mm In the second case with a 200 mm air column, the height of mercury is 746 mm: Again, applying the formula: P_0 (200 mm) = 746 mm + 200 mm = 946 mm Setting these two equations equal to atmospheric pressure, we have: P_atm = P_0 (250 mm) = P_0 (200 mm) + 4 mm, Substituting and manipulating gives us: P_atm = (1000 mm + 946 mm)/2 = 973 mm. Now, converting the height of mercury into pressure in pascals: P_atm = 973 mm * 13,600 Pa/mm = 13,235,000 Pa or 13.2 kPa (approximately) Thus, the atmospheric pressure is around 973 mm of mercury, or about 13.2 kPa.

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