What is the temperature of air, given P= 90 kPA and the pressure is 1.0 kg(m^-3)?
Thanks!
To determine the temperature of air, we need to use the ideal gas law, which states that:
PV = nRT
where:
P = pressure (in Pascal)
V = volume (in cubic meters)
n = number of moles of gas
R = ideal gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
However, in the given information, only the pressure and the density of air are provided. Therefore, we first need to calculate the volume.
To find the volume (V), we can use the relationship between density (ρ) and mass (m):
ρ = m/V
where:
ρ = density (in kg/m^3)
m = mass of air (in kg)
V = volume (in cubic meters)
Since the density is given as 1.0 kg/m^3, and we know that density is mass divided by volume, we can rearrange the equation to solve for volume:
V = m/ρ
Now, to calculate the mass (m), we can use the relationship between mass, density, and volume:
m = ρV
Substituting the values, we have:
m = (1.0 kg/m^3) × V
Next, we can substitute the calculated values for V and m into the ideal gas law equation:
PV = nRT
(90 kPa) = (1.0 kg/m^3) × V × R × T
Since the pressure is given in kilopascal (kPa), we need to convert it to Pascal (Pa). 1 kPa is equal to 1000 Pa. Therefore, we have:
(90,000 Pa) = (1.0 kg/m^3) × V × (8.314 J/(mol·K)) × T
Let's assume that we have 1 mole of air, so we can simplify the equation to:
(90,000 Pa) = V × (8.314 J/K) × T
We can rearrange this equation to solve for temperature (T):
T = (90,000 Pa) / [V × (8.314 J/K)]
Now that we have the formula, we can substitute the calculated value for V to find the temperature (T).