Problem 57 - University Physics with Modern Physics

Question

Atmosphere of Titan. Titan, the largest satellite of Saturn, has a thick nitrogen atmosphere. At its surface, the pressure is 1.5 Earth-atmospheres and the temperature is 94 \(\mathrm{K}\) . (a) What is the surface temperature in \(^{\circ} \mathrm{C}\) ? (b) Calculate the surface density in Titar's atmosphere in molecules per cubic meter (c) Compare the density of Titan's surface atmosphere to the density of Earth's atmosphere at \(22^{\circ} \mathrm{C}\) . Which body has denser atmosphere?

Step-by-Step Solution

Answer

(a) \(-179.15 \,^{\circ}\text{C}\), (b) \(1.19 \times 10^{26}\text{ molecules/m}^3\), (c) Titan's atmosphere is denser.

Step 1: Convert temperature to Celsius

We start with the surface temperature given in Kelvin, which is 94 K. To convert it to degrees Celsius, use the formula: \[ T_{\text{C}} = T_{\text{K}} - 273.15 \]Substitute 94 for \( T_{\text{K}} \):\[ T_{\text{C}} = 94 - 273.15 = -179.15 \,^{\circ}\text{C} \]

Step 2: Find surface density in molecules per cubic meter

We apply the ideal gas law, which is given by:\[ PV = nRT \] where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. We rearrange it to find density, \( \rho \), using \( \rho = \frac{nM}{V} = \frac{PM}{RT} \), where M is the molar mass of nitrogen (28.02 kg/kmol. Given \( P = 1.5 \) atm = 152 kPa after converting (1 atm = 101.3 kPa), \( T = 94 \) K, and \( R = 8.314 \, J/(mol \cdot K) \). The density \( \rho \) in kg/m^3 is:\[ \rho = \frac{152 \times 28.02}{8.314 \times 94} \approx 5.51 \, \text{kg/m}^3 \] Each nitrogen molecule has a mass of approximately \( 4.65 \times 10^{-26} \) kg, thus density in molecules/m³ is:\[ \rho = \frac{5.511}{4.65 \times 10^{-26}} \approx 1.19 \times 10^{26} \, \text{molecules/m}^3 \]

Step 3: Compare with Earth's atmosphere density

To compare, first calculate Earth's atmosphere density at \( 22^{\circ}\text{C} \). Convert Celsius to Kelvin: \[ T_{\text{K}} = 22 + 273.15 = 295.15 \,K \]Approximate density of Earth's atmosphere using ideal gas law is known to be about 1.2 kg/m³. With similar calculations: \[ \rho_{\text{Earth}} = \frac{101.3 \times 28.97}{8.314 \times 295.15} \approx 1.2 \, \text{kg/m}^3 \] Since Titan's density (5.51 kg/m³) is much higher than Earth's (1.2 kg/m³), Titan's atmosphere is denser.

Key Concepts

Titan Atmosphere

Titan, Saturn's biggest moon, catches interest due to its thick, nitrogen-rich atmosphere. Unlike any other moon, Titan's atmosphere is even denser than Earth's.

• The surface pressure is measured at about 1.5 times that of Earth's, indicating a substantial gaseous envelop.
• Nitrogen dominates Titan's atmosphere composition, similar to Earth but much frostier.

Given these challenges, Titan presents an intriguing case study for atmospheric sciences, drawing comparisons with Earth to understand planetary atmospheres better.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in understanding how gases behave under various conditions. It is formally expressed as:\[ PV = nRT \]where:- \( P \) is the pressure,- \( V \) is the volume,- \( n \) is the number of moles,- \( R \) is the ideal gas constant, and- \( T \) is the temperature in Kelvin.In essence, this equation allows scientists to predict one state of a gas if the others are known. On Titan, by knowing its atmospheric pressure and surface temperature, one can estimate the number of gas molecules present.

Surface Pressure

Surface pressure provides insight into the thickness and density of an atmosphere. Titan's surface pressure is 1.5 times that of Earth's standard atmospheric pressure.

• Pressurized atmospheres like these imply a heavier load of gases above the surface.
• This leads to denser, more substantial atmospheric compositions compared to less pressurized bodies.

Understanding surface pressure is critical not only in weather prediction but also in comprehending the environmental conditions and potential habitability of celestial bodies.

Temperature Conversion

Temperature conversion is essential when dealing with scientific data. In this context, converting from Kelvin to Celsius is routine for understanding temperatures in a human context. The conversion formula from Kelvin to Celsius is:\[ T_{C} = T_{K} - 273.15 \]Using this equation, Titan's surface temperature of 94 Kelvin converts to:\[ T_{C} = 94 - 273.15 = -179.15 \,^{\circ}C \]Such extreme temperatures highlight the frigid conditions on Titan, which influence its atmospheric behavior and chemical processes.