91Ó°ÊÓ

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 \( T_{\text{K}} \):\[ T_{\text{C}} = 94 - 273.15 = -179.15 \,^{\circ}\text{C} \]

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:\[ \rho = \frac{5.511}{4.65 \times 10^{-26}} \approx 1.19 \times 10^{26} \, \text{molecules/m}^3 \]

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.