Question

If the air is contained in a tyre at 2.0 atm gauge pressure, then absolute pressure of air will be

• A. \(1.013 \times 10^5 \text{ Pa}\)
• B. \(2.026 \times 10^5 \text{ Pa}\)
• C. \(3.04 \times 10^5 \text{ Pa}\)
• D. \(4.052 \times 10^5 \text{ Pa}\)

Hint:

To simplify calculations, add the pressures in atmospheric units (\(\text{atm}\)) first before converting to Pascals:
\[ P_{\text{abs}} = 2.0 \text{ atm (gauge)} + 1.0 \text{ atm (atmosphere)} = 3.0 \text{ atm (absolute)} \] Then: \(3.0 \times 1.013 \times 10^5 \text{ Pa} = 3.039 \times 10^5 \text{ Pa}\). This minimizes arithmetic errors.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Pressure measurements are defined based on their reference point. Gauge pressure is measured relative to local atmospheric pressure, while absolute pressure is measured relative to a perfect vacuum (absolute zero pressure).
Key Formula or Approach:
The relationship between absolute pressure (\(P_{\text{abs}}\)) and gauge pressure (\(P_{\text{gauge}}\)) is given by: \[ P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}} \] Standard atmospheric pressure is defined as: \[ 1 \text{ atm} \approx 1.013 \times 10^5 \text{ Pa} \]

Step 2: Detailed Explanation:
We are given: \[ P_{\text{gauge}} = 2.0 \text{ atm} \] Convert the gauge pressure into Pascals (\(\text{Pa}\)): \[ P_{\text{gauge}} = 2.0 \times 1.013 \times 10^5 \text{ Pa} = 2.026 \times 10^5 \text{ Pa} \] The standard atmospheric pressure at sea level is: \[ P_{\text{atm}} = 1 \text{ atm} = 1.013 \times 10^5 \text{ Pa} \] Now, calculate the absolute pressure: \[ P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}} \] \[ P_{\text{abs}} = 2.026 \times 10^5 \text{ Pa} + 1.013 \times 10^5 \text{ Pa} \] \[ P_{\text{abs}} = 3.039 \times 10^5 \text{ Pa} \] Rounding to three significant figures yields: \[ P_{\text{abs}} \approx 3.04 \times 10^5 \text{ Pa} \] Thus, the absolute pressure of the air inside the tyre is \(3.04 \times 10^5 \text{ Pa}\).

Step 3: Final Answer:
The absolute pressure is \(3.04 \times 10^5 \text{ Pa}\), which corresponds to Option (C).