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Chapter 10: Problem 22

(a) On Titan, the largest moon of Saturn, the atmospheric pressure is \(1.63105 \mathrm{~Pa}\). What is the atmospheric pressure of Titan in atm? (b) On Venus the surface atmospheric pressure is about 90 Earth atmospheres. What is the Venusian atmospheric pressure in kilopascals?

Short Answer

Expert verified
(a) The atmospheric pressure of Titan is approximately \(1.61 \times 10^{-5}\) atm. (b) The Venusian atmospheric pressure is approximately 9119.25 kPa.

Step by step solution

01

Convert the atmospheric pressure of Titan from Pa to atm

To convert the given pressure of Titan (1.63105 Pa) to atm, we will use the conversion factor for atmospheres to Pascals: 1 atm = 101325 Pa Now, divide the given pressure in Pa by the conversion factor: Titan's pressure in atm = \( \frac{1.63105~Pa}{101325~Pa/atm} \)
02

Calculate Titan's pressure in atm

Now let's perform the division to find the atmospheric pressure of Titan in atm: Titan's pressure in atm = \( \frac{1.63105}{101325} \) = 1.61 x 10^{-5} atm So, the atmospheric pressure of Titan is approximately 1.61 x 10^{-5} atm.
03

Convert the atmospheric pressure of Venus from Earth atmospheres to kPa

To convert the given pressure of Venus (90 Earth atmospheres) to kPa, we will use the conversion factor for atmospheres to kilopascals: 1 atm = 101.325 kPa Now, multiply the given pressure in Earth atmospheres by the conversion factor: Venus's pressure in kPa = 90 atm x 101.325 kPa/atm
04

Calculate Venus's pressure in kPa

Now let's perform the multiplication to find the atmospheric pressure of Venus in kPa: Venus's pressure in kPa = 90 x 101.325 = 9119.25 kPa So, the atmospheric pressure of Venus is approximately 9119.25 kPa.

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Key Concepts

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

Pressure Unit Conversion
Understanding pressure unit conversion is essential in fields such as meteorology, aviation, and engineering. Pressure, which is force applied per unit area, can be expressed in various units, with Pascals (Pa), atmospheres (atm), and kilopascals (kPa) being among the most commonly used.

Converting between these units requires familiarity with their equivalence: 1 atm is equal to 101325 Pa and also to 101.325 kPa. The process of conversion typically involves multiplication or division, depending on whether you are converting to a larger or smaller unit of measure. This knowledge allows students to seamlessly switch between units to interpret data or solve problems that are given in a different pressure scale.
Pascals to Atmospheres
The Pascal (Pa) is a standard unit of pressure in the International System of Units (SI) and is defined as one newton per square meter. However, when dealing with atmospheric pressure, using atmospheres (atm) is more intuitive as 1 atm denotes the average atmospheric pressure at sea level.

To convert Pascals to atmospheres, divide the pressure in Pascals by 101325, the value of one atmosphere in Pascals. For example, the conversion from the given problem shows that the atmospheric pressure of Titan \(1.63105~\text{Pa}\) is approximately \(1.61 \times 10^{-5}\) atm when applying the conversion factor.
Atmospheres to Kilopascals
Though 'atmosphere' is a unit that helps relate pressure terms to everyday experiences, kilopascals (kPa) provide a bridge between the common atmospheric scale and the precise measurements used in scientific research. To convert atmospheres to kilopascals, you simply multiply by 101.325, the number of kilopascals in one atmosphere.

In practice, to convert the pressure on Venus from Earth atmospheres to kilopascals, you would take the given 90 atm and multiply by the conversion factor, resulting in \(9119.25~\text{kPa}\). This step is critical for ensuring measurements are in the right scale for the application at hand, such as evaluating the structural integrity of spacecraft materials or atmospheric studies.

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Most popular questions from this chapter

Large amounts of nitrogen gas are used in the manufacture of ammonia, principally for use in fertilizers. Suppose \(120.00 \mathrm{~kg}\) of \(\mathrm{N}_{2}(g)\) is stored in a 1100.0-L metal cylinder at \(280^{\circ} \mathrm{C}\). (a) Calculate the pressure of the gas, assuming ideal-gas behavior. (b) By using data in Table 10.3, calculate the pressure of the gas according to the van der Waals equation. (c) Under the conditions of this problem, which correction dominates, the one for finite volume of gas molecules or the one for attractive interactions?

Calculate the number of molecules in a deep breath of air whose volume is \(2.25 \mathrm{~L}\) at body temperature, \(37^{\circ} \mathrm{C}\), and a pressure of 735 torr.

How does a gas differ from a liquid with respect to each of the following properties: (a) density, (b) compressibility, (c) ability to mix with other substances of the same phase to form homogeneous mixtures?

(a) Place the following gases in order of increasing average molecular speed at \(25^{\circ} \mathrm{C}: \mathrm{Ne}, \mathrm{HBr}, \mathrm{SO}_{2}, \mathrm{NF}_{3}, \mathrm{CO}\). (b) Calculate the rms speed of \(\mathrm{NF}_{3}\) molecules at \(25^{\circ} \mathrm{C}\).

(a) How high in meters must a column of water be to exert a pressure equal to that of a 760 -mm column of mercury? The density of water is \(1.0 \mathrm{~g} / \mathrm{mL}\), whereas that of mercury is \(13.6 \mathrm{~g} / \mathrm{mL}\). (b) What is the pressure in atmospheres on the body of a diver if he is \(39 \mathrm{ft}\) below the surface of the water when atmospheric pressure at the surface is \(0.97\) atm?
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