Question

A local weather station reports the barometric pressure as \(74.9 \mathrm{cmHg}\) (centimeters of \(\mathrm{Hg}\) ). Convert this pressure to torr and to atm.

Short answer

Pressure: 74.9 torr; approximates to 0.986 atm.

Step-by-step solution

Understanding Unit Conversion

The problem asks us to convert barometric pressure from centimeters of mercury (cmHg) to torr and then to atmospheres (atm). The relationship between cmHg and torr is 1:1, meaning 1 cmHg equals 1 torr. For conversion to atm, knowing that 1 atm equals 76 cmHg is useful.

Conversion from cmHg to Torr

Since 1 cmHg is equal to 1 torr, the pressure in torr is directly the same as in cmHg. So, the conversion is simple: \[ 74.9\, \text{cmHg} = 74.9\, \text{torr} \]

Conversion from Torr to Atmospheres

To convert from torr to atm, use the relationship that 1 atm = 76 torr. We set up the conversion as follows: \[\text{Pressure in atm} = \frac{74.9 \text{ torr}}{76 \text{ torr/atm}}\]This simplifies to: \[\text{Pressure in atm} \approx 0.9855 \text{ atm}\]

Final Verification

The final pressures, 74.9 torr and approximately 0.986 atm, should be verified to ensure they are reasonable conversions based on known values and relationships.

Key concepts

barometric pressure

Barometric pressure is the pressure exerted by the atmosphere at a given point. It is due to the weight of the air above us and can vary depending on weather conditions and altitude. Instruments called barometers measure this type of pressure, hence the name "barometric pressure." Understanding this concept is important because barometric pressure influences weather patterns and altitude adjustments as it changes. For students interested in meteorology or physics, grasping how barometric pressure is measured and why it matters can be quite fascinating. Measurements of barometric pressure are taken using units such as centimeters of mercury (cmHg), inches of mercury (inHg), or millibars (mb).

• Barometric pressure helps predict weather changes.
• It is a key component in understanding atmospheric conditions.
• Higher altitudes generally have lower pressures due to less air above.

Knowing the unit of measurement used for barometric pressure, such as cmHg, is essential for accurate conversions and interpretations.

torr to atm conversion

Converting between torr and atmospheres (atm) is a vital skill in scientific studies, especially in fields like physics and chemistry. The torr is a unit of pressure named after the Italian physicist Evangelista Torricelli, an important figure in fluid dynamics.For students, it's helpful to understand that:

• Standard atmospheric pressure is defined as 1 atm, which is equivalent to 760 torr.
• Thus, to convert a pressure reading from torr to atm, you simply divide the pressure in torr by 760. This is because 1 atm = 760 torr.
• In our problem, converting 74.9 torr to atm involves: \[ \text{Pressure in atm} = \frac{74.9 \text{ torr}}{760 \text{ torr/atm}} \approx 0.9855 \text{ atm} \]

This conversion is straightforward but essential for understanding pressure in terms everyone in the scientific community uses.

centimeters of mercury (cmHg)

The use of centimeters of mercury (cmHg) as a unit of measurement roots back to traditional methods of measuring pressure using a mercury column. One cmHg represents the pressure exerted by a column of mercury one centimeter high under the force of gravity. It is a convenient and direct way to express pressure, especially in weather and laboratory settings. Students should note that:

• 1 cmHg is equivalent to 1 torr, making some conversions between them direct and easy.
• The relationship stems from historical and experimental conditions, where 1 cmHg was defined to equal 1 torr.
• This simplifies many calculations without losing accuracy.

For practical computations, converting pressure from cmHg to atm follows another simple step. Knowing that 1 atm equals 76 cmHg enables easy conversion, crucial for accurately presenting scientific data across various unit systems. Mastery of these conversions aids in better understanding other pressure-related applications in science and engineering.