To determine the density of the combusted air-fuel mixture, we can use the ideal gas law, which states:
\[ PV = nRT \]
Rearranging this for density (\(\rho\)), we can express it in terms of pressure (\(P\)), the gas constant (\(R\)), and temperature (\(T\)):
\[ \rho = \frac{P}{RT} \]
Step 1: Convert the temperature from Celsius to Kelvin
The temperature in Kelvin (\(T\)) is given by:
\[ T(K) = T(°C) + 273.15 \]
Substituting \(T = 582\) °C:
\[ T(K) = 582 + 273.15 = 855.15 , \text{K} \]
Step 2: Convert the pressure from bar to Pascals
Pressure in pascals is required for the calculation. Since \(1 , \text{bar} = 100,000 , \text{Pa}\):
\[ P = 6.2 , \text{bar} = 6.2 \times 100,000 , \text{Pa} = 620,000 , \text{Pa} \]
Step 3: Substitute values into the density equation
Now we can substitute \(P\), \(R\), and \(T\) into the density equation:
\[ \rho = \frac{P}{RT} \]
Substituting in:
\(P = 620,000 , \text{Pa}\),
\(R = 287.00 , \text{J/(kgK)}\),
\(T = 855.15 , \text{K}\)
\[ \rho = \frac{620,000}{287.00 \times 855.15} \]
Calculating the denominator:
\[ 287.00 \times 855.15 \approx 245,792.55 \]
Now, calculating the density \(\rho\):
\[ \rho = \frac{620,000}{245,792.55} \approx 2.52 , \text{kg/m}^3 \]
Final Answer
The density of the combusted air-fuel mixture is approximately:
\[ \mathbf{2.52 , kg/m^3} \]