Consider air to be a mixture of 80% nitrogen and 20% oxygen.
How much work is required to raise the temperature of 7 mol of air by 9 ∘C in an adiabatic process?
To calculate the work required to raise the temperature of air in an adiabatic process, we need to use the equation for adiabatic work:
W = -Cv * (Tf - Ti)
Where:
W is the work done on the air (in this case, since the temperature is increasing, work is done on the air),
Cv is the molar heat capacity at constant volume of the air,
Tf is the final temperature, and
Ti is the initial temperature.
To find the molar heat capacity at constant volume (Cv), we can use the molar heat capacities of nitrogen (N2) and oxygen (O2) gas, and their respective mole fractions in air.
The molar heat capacities at constant volume (Cv) for N2 and O2 are approximately 20.8 J/(mol·K) and 21.1 J/(mol·K), respectively.
Given that air is composed of 80% nitrogen and 20% oxygen, we can calculate the molar heat capacity of air by averaging the molar heat capacities of N2 and O2 weighted by their mole fractions:
Cv_air = (0.8 * Cv_N2) + (0.2 * Cv_O2)
Let's calculate Cv_air:
Cv_N2 = 20.8 J/(mol·K)
Cv_O2 = 21.1 J/(mol·K)
Cv_air = (0.8 * 20.8) + (0.2 * 21.1)
= 20.8 J/(mol·K)
Now, we can substitute the values into the formula for work:
W = -Cv_air * (Tf - Ti)
Given:
Tf = Initial temperature + change in temperature
= Ti + 9 ∘C
= Ti + 9 K (since the temperature is in Kelvin)
Substituting the values into the work equation:
W = -Cv_air * (Tf - Ti)
= -20.8 J/(mol·K) * (Ti + 9 K - Ti)
= -20.8 J/mol * 9 K
= -187.2 J/mol
Therefore, the work required to raise the temperature of 7 mol of air by 9 ∘C in an adiabatic process is -187.2 J/mol. Note that the negative sign indicates work done on the system.