Answer
We can use the ideal gas law to calculate the volume occupied by 1.00 mol of Titan's atmosphere:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.
First, we need to convert the temperature from Celsius to Kelvin:
T = -180°C + 273.15 = 93.15 K
Next, we can use the ideal gas law to solve for the volume of 1.00 mol of gas:
V = nRT/P
We know that P = 1.47 atm (since "standard atmospheric pressure" on Earth is defined as 1 atm), n = 1.00 mol, R = 0.0821 L·atm/(mol·K), and T = 93.15 K. Substituting these values, we get:
V = (1.00 mol)(0.0821 L·atm/(mol·K))(93.15 K)/(1.47 atm) = 4.18 L
So 1.00 mol of Titan's atmosphere occupies a volume of 4.18 L under the given conditions of pressure and temperature.
Work Step by Step
We can use the ideal gas law to calculate the volume occupied by 1.00 mol of Titan's atmosphere:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.
First, we need to convert the temperature from Celsius to Kelvin:
T = -180°C + 273.15 = 93.15 K
Next, we can use the ideal gas law to solve for the volume of 1.00 mol of gas:
V = nRT/P
We know that P = 1.47 atm (since "standard atmospheric pressure" on Earth is defined as 1 atm), n = 1.00 mol, R = 0.0821 L·atm/(mol·K), and T = 93.15 K. Substituting these values, we get:
V = (1.00 mol)(0.0821 L·atm/(mol·K))(93.15 K)/(1.47 atm) = 4.18 L
So 1.00 mol of Titan's atmosphere occupies a volume of 4.18 L under the given conditions of pressure and temperature.
