Final answer:
To find the density of an ideal gas at STP when given its density at different conditions, we use the ideal gas law relationship that the density is directly proportional to pressure and inversely proportional to temperature. By properly adjusting initial conditions to STP conditions, the new density of the ideal gas can be calculated to be approximately 0.446 g/L.
Step-by-step explanation:
If the density of an ideal gas is 4.472 g/L at 126.89 °C and 124.09 kPa, we can use the ideal gas law and the concept of STP to find its density at Standard Temperature and Pressure (STP). To solve this problem, we apply the ideal gas law (PV = nRT) indirectly through a derived formula that relates the density (ρ), molar mass (M), pressure (P), and temperature (T): ρ = (PM) / (RT).
Given that STP is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm (101.3 kPa), we need to adjust from the initial conditions (4.472 g/L at 126.89 °C and 124.09 kPa) to STP. We can use the fact that density is directly proportional to pressure and inversely proportional to temperature (in Kelvin) to find the new density.
The calculation will be based on converting the initial pressure and temperature to their STP equivalents using the ideal gas law, keeping in mind that the number of moles and the volume are constant.
First, we convert the initial temperature to Kelvin and the pressure to atmospheres:
T1 = 126.89°C = 400.04 K
P1 = 124.09 kPa = 1.225 atm
Now we can write the proportional relationship: ρ₂ / ρ₁ = (P₂ / P₁) × (T₁ / T₂)
Applying the values: ρ₂ / 4.472 g/L = (1 atm / 1.225 atm) × (400.04 K / 273.15 K)
Solving for ρ₂ gives us: ρ₂ = ((1 / 1.225) × (400.04 / 273.15)) × 4.472 g/L
After calculating this, the new density, ρ₂, comes out to be approximately 0.446 g/L.
Hence, the correct answer to the question is option a. 0.446 g/L.