The pressure 45.0 m under water is 543 kPa. What is this pressure in atmospheres (atm)? P atm What is this pressure in millimeters of mercury (mmHg)? PE mmHg
The Correct Answer And Explanation is:
To convert the pressure 543 kPa into atmospheres (atm) and millimeters of mercury (mmHg), we use the following conversion factors:
- 1 atm = 101.325 kPa
- 1 atm = 760 mmHg
✅ Step 1: Convert kPa to atm
Pressure in atm=543 kPa101.325 kPa/atm≈5.36 atm\text{Pressure in atm} = \frac{543 \, \text{kPa}}{101.325 \, \text{kPa/atm}} \approx 5.36 \, \text{atm}
✅ Step 2: Convert atm to mmHg
Pressure in mmHg=5.36 atm×760 mmHg/atm≈4074 mmHg\text{Pressure in mmHg} = 5.36 \, \text{atm} \times 760 \, \text{mmHg/atm} \approx 4074 \, \text{mmHg}
✅ Final Answers:
- 5.36 atm
- 4074 mmHg
🧠 300-Word Explanation:
Pressure is a measure of force applied over an area. In this question, we are given pressure at a depth of 45.0 meters under water as 543 kilopascals (kPa) and are asked to convert this pressure into two other common units: atmospheres (atm) and millimeters of mercury (mmHg).
First, let’s understand what each unit represents:
- Kilopascal (kPa) is a metric unit of pressure. The Pascal (Pa) is the SI unit, and 1 kPa = 1,000 Pa.
- Atmosphere (atm) is a unit based on average atmospheric pressure at sea level. By definition, 1 atm = 101.325 kPa.
- Millimeters of mercury (mmHg) is a unit based on the height of a column of mercury that pressure can support. 1 atm = 760 mmHg.
To convert from kPa to atm, divide the given pressure in kPa by the conversion factor (101.325). This gives you the equivalent value in atmospheres. Since 543 kPa is significantly higher than the atmospheric pressure at sea level (101.325 kPa), it makes sense that the answer is more than 1 atm — specifically, about 5.36 atm.
Next, to convert from atm to mmHg, you multiply the result by 760. So, 5.36 atm × 760 mmHg/atm = 4074 mmHg, which is also a large number and again expected due to the depth under water.
These conversions are useful in physics, chemistry, and engineering, especially when comparing pressures across different systems or understanding how environmental conditions (like depth) affect pressure.