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In meteorology pressures are expressed in units of millibars (mbar). Convert 985 mbar to torr and atmospheres.

Short Answer

Expert verified
985 mbar is approximately 738.56 torr and 0.971 atm.

Step by step solution

01

Introduction to Conversion Factors

In this exercise, we need to convert millibars to two different units: torr and atmosphere. To do this, we use the following conversion factors: 1 mbar = 0.750062 torr and 1 mbar = 0.000986923 atm.
02

Convert Millibars to Torr

To convert 985 mbar to torr, multiply the pressure in millibars by the conversion factor from mbar to torr. \( 985 ext{ mbar} \times 0.750062 \text{ torr/mbar} = 738.56097 \text{ torr} \)Therefore, 985 mbar is equivalent to approximately 738.56 torr.
03

Convert Millibars to Atmospheres

Similarly, to convert 985 mbar to atmospheres, multiply the pressure in millibars by the conversion factor from mbar to atm.\( 985 ext{ mbar} \times 0.000986923 \text{ atm/mbar} = 0.971311 \text{ atm} \)Hence, 985 mbar is equivalent to approximately 0.971 atm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Millibar
A millibar (mbar) is a metric unit of pressure used widely in meteorology. This unit is helpful for measuring atmospheric pressure which influences weather patterns. Understanding millibars is crucial because pressure measurements in this unit are commonly used in weather reports. For example, a standard atmospheric pressure at sea level is about 1013.25 mbar. To comprehend how millibars work in real-world applications, consider that when meteorologists talk about areas of high or low pressure, they often refer to them in millibars. This allows them to communicate effectively about weather conditions such as storms or calm weather. If you're curious about converting millibars to other units, it's essential to know the conversion factors mentioned in your original textbook exercise. By using these factors, you can easily convert millibars to other common units of pressure.
Torr
The torr is another unit of pressure, primarily used in fields like physics and engineering. Named after Evangelista Torricelli, an Italian scientist who invented the barometer, "torr" quantifies pressure quite differently from millibars. Another useful detail about the torr is its relationship with millimeters of mercury (mmHg); 1 torr is almost equivalent to 1 mmHg. The conversion between millibars and torr involves knowing that 1 millibar equals approximately 0.750062 torr. This conversion is helpful when you need to switch between units for scientific purposes or for analyzing various data types. To apply this knowledge in practice, in the example provided, 985 mbar was converted into torr resulting in approximately 738.56 torr. This conversion is crucial because it allows scientists and meteorologists to collate and compare data logged in different units, seamlessly bridging European and American data traditions.
Atmospheres
An atmosphere (atm) is a unit of pressure based on the average atmospheric pressure at sea level on Earth. It provides a more "general" way to express pressure that is easily relatable. Using atmospheres is quite common in chemistry and physics, particularly when dealing with gases. This unit allows scientists to conceptualize how gases behave and interact under different pressures. For converting millibars into atmospheres, the exercise illustrates that 1 millibar equals approximately 0.000986923 atm. This means when you convert 985 mbar, it equates to roughly 0.971 atm. By understanding this conversion, students can better grasp how changes in atmospheric pressure can affect gas volumes, for example. It's a critical conversion for fields where understanding gas laws is crucial, such as chemistry and atmospheric sciences.

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Most popular questions from this chapter

A gaseous mixture contains \(400.0\) Torr of \(\mathrm{H}_{2}(g), 355.1\) Torr of \(\mathrm{N}_{2}(g)\), and \(75.2\) Torr of \(\operatorname{Ar}(g)\). Calculate the mole fraction of each of these gases.

Calculate the volume of \(0.200 \mathrm{M} \mathrm{NaOH}\) required to prepare 150 milliliters of \(\mathrm{H}_{2}(g)\) at \(10.0^{\circ} \mathrm{C}\) and 750 Torr from the reaction described by the equation \(2 \mathrm{Al}(s)+2 \mathrm{NaOH}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow\) \(2 \mathrm{NaAlO}_{2}(a q)+3 \mathrm{H}_{2}(g)\)

A balloon has a volume of \(1.2\) liters on a warm \(32^{\circ} \mathrm{C}\) day. If the same balloon is placed in the freezer and cooled to \(-18^{\circ} \mathrm{C}\), what volume will it occupy? (Assume constant pressure.)

Sulfur dioxide can combine with oxygen to form sulfur trioxide according to the equation $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g) $$ A researcher studying the reaction under various conditions introduces \(725.0\) Pa of sulfur dioxide into a rigid stainless steel reaction chamber maintained at a constant temperature. She then introduces \(500.0 \mathrm{~Pa}\) of oxygen. Assuming that all the sulfur dioxide is converted to sulfur trioxide, calculate the final total pressure inside the chamber.

A gaseous mixture inside a rigid steel vessel contains \(75 \% \mathrm{CO}_{2}(g)\) and \(25 \% \mathrm{H}_{2} \mathrm{O}(g)\) by volume at \(175^{\circ} \mathrm{C}\) and \(225 \mathrm{kPa}\). The mixture is then cooled to \(0^{\circ} \mathrm{C}\), thereby condensing the water vapor. What is the pressure of the \(\mathrm{CO}_{2}(g)\) at \(0^{\circ} \mathrm{C} ?\) (Assume that there is no water vapor present after condensation.)

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