A manometer connected to a sealed box shows a mercury column height difference of 0.250 m. Given the density of mercury as 13600 kg/m^3 and assuming atmospheric pressure is 1.00 × 10^5 Pa, what is the gas pressure inside the box in pascals?

• A. 1.00 × 10^5 Pa
• B. 1.03 × 10^5 Pa
• C. 1.36 × 10^5 Pa
• D. 9.66 × 10^4 Pa

Answer: (B)

Verified step by step guidance

1. Identify the type of manometer and the pressure relationship it represents. Since the mercury column height difference is given, the pressure inside the box is either higher or lower than atmospheric pressure by the pressure exerted by the mercury column.
2. Calculate the pressure exerted by the mercury column using the hydrostatic pressure formula: \(P = \rho g h\), where \(\rho\) is the density of mercury (13600 kg/m^3), \(g\) is the acceleration due to gravity (9.8 m/s^2), and \(h\) is the height difference (0.250 m).
3. Determine whether the gas pressure inside the box is greater or less than atmospheric pressure by analyzing the manometer setup (e.g., if mercury is pushed down on the side open to atmosphere, gas pressure is higher; if mercury is pushed down on the side connected to the gas, gas pressure is lower).
4. Calculate the gas pressure inside the box by adding or subtracting the mercury pressure from atmospheric pressure: \(P_{gas} = P_{atm} \pm \rho g h\), depending on the direction determined in the previous step.
5. Express the final gas pressure in pascals (Pa) using the values and units consistently, ensuring the answer matches the expected order of magnitude.