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 find the amount of work required in an adiabatic process, we need to use the first law of thermodynamics, which states that the change in internal energy (∆U) is equal to the heat added (Q) minus the work done (W).
Since the process is adiabatic, it means that no heat is exchanged with the surroundings, so the heat added (Q) is zero.
Therefore, the equation becomes: ∆U = -W
To calculate the change in internal energy (∆U), we use the equation: ∆U = n*Cv*∆T, where n is the number of moles, Cv is the molar specific heat at constant volume, and ∆T is the change in temperature.
Since air is a mixture of 80% nitrogen and 20% oxygen, we can assume the molar specific heat at constant volume (Cv) for air is the average of the Cv values for nitrogen and oxygen.
The molar specific heat at constant volume for nitrogen (N2) is approximately 20.8 J/(mol*K), and for oxygen (O2) is approximately 20.8 J/(mol*K). Thus, we can assume the molar specific heat at constant volume for air is approximately 20.8 J/(mol*K).
Now, let's calculate ∆U:
∆U = n*Cv*∆T
∆U = 7 mol * (20.8 J/(mol*K)) * 9 ∘C
∆U = 1310.4 J
Since ∆U is negative because the internal energy decreases during the adiabatic process, we have:
-∆U = W
W = -(-1310.4 J)
W = 1310.4 J
Therefore, the work required to raise the temperature of 7 mol of air by 9 ∘C in an adiabatic process is approximately 1310.4 J.