Use the PV diagram of a system. The units for P and V are atmospheric pressure atm and liter L. Process A to B and B to C are straight lines on it.
How much is the work done by the system from A to C in J?
Given these coordinates:(volume on the x, pressure on the y)
A: (1 Li, 3.0 atm)
B: (5 Li, 6.0 atm)
C: (11 Li, 4.5 atm)
A: p₁=3 atm =3.04•10⁵ Pa,
V₁=1 L = 10⁻³ m³;
B: p₂=6 atm = 6.08•10⁵ Pa,
V₂ =5L=5•10⁻³m³;
C: p₃= 4.5 atm = 4.56•10⁵ Pa,
V₃=11L = 11•10⁻³m³.
A->B:
W₁= 0.5(p₁+p₂)( V₂-V₁) =
=0.5(3.04+6.08)•10⁵•(5-1)•10⁻³ =
=1824 J
B->C:
W₂ =0.5(p₂ +p₃)( V₃-V₂) =
=0.5( 6.08+4.56)•10⁵•(11-5)•10⁻³ =
=3192 J
W= W₁+W₂ =1824 + 3192 =5016 J
To calculate the work done by the system from point A to point C, we need to use the PV diagram and the formula for work in a thermodynamic system.
The formula for work in a thermodynamic system is given by:
Work = ∫PdV
Where P is the pressure and V is the volume.
First, we need to determine the equation of the line for process A to B. We can use the two given points, A (1 Li, 3.0 atm) and B (5 Li, 6.0 atm), to find the equation of the line.
The slope of the line can be calculated using the formula:
slope = (change in pressure) / (change in volume)
slope = (6.0 atm - 3.0 atm) / (5 Li - 1 Li)
slope = 3.0 atm / 4 Li
slope = 0.75 atm/Li
Now, using the equation of a line (y = mx + b), we can solve for the value of b:
b = y - mx
b = 3.0 atm - (0.75 atm/Li × 1 Li)
b = 2.25 atm
So, the equation for process A to B is:
P = 0.75 V + 2.25
Similarly, we can determine the equation for process B to C using the given points B (5 Li, 6.0 atm) and C (11 Li, 4.5 atm).
slope = (4.5 atm - 6.0 atm) / (11 Li - 5 Li)
slope = -1.5 atm / 6 Li
slope = -0.25 atm/Li
b = y - mx
b = 6.0 atm - (-0.25 atm/Li × 5 Li)
b = 7.25 atm
So, the equation for process B to C is:
P = -0.25 V + 7.25
Now that we have the equations for both processes, we can calculate the work done from A to C by finding the area under the curve between the two processes.
The work done from A to C can be calculated as the sum of the work done in process A to B and process B to C:
Work = Work(A to B) + Work(B to C)
To calculate the work done in each process, we need to integrate the respective equations over their corresponding volume ranges. For process A to B, the volume ranges from 1 Li to 5 Li, and for process B to C, the volume ranges from 5 Li to 11 Li.
Let's calculate the work done in process A to B:
Work(A to B) = ∫PdV (from 1 Li to 5 Li)
Work(A to B) = ∫(0.75 V + 2.25)dV (from 1 Li to 5 Li)
Work(A to B) = ∫(0.75 V)dV + ∫(2.25)dV (from 1 Li to 5 Li)
Work(A to B) = (0.75/2)V^2 + 2.25V (from 1 Li to 5 Li)
Plugging in the volume values:
Work(A to B) = (0.75/2)(5^2 - 1^2) + 2.25(5 - 1)
Now calculate the work done in process B to C:
Work(B to C) = ∫PdV (from 5 Li to 11 Li)
Work(B to C) = ∫(-0.25 V + 7.25)dV (from 5 Li to 11 Li)
Work(B to C) = ∫(-0.25 V)dV + ∫(7.25)dV (from 5 Li to 11 Li)
Work(B to C) = (-0.25/2)V^2 + 7.25V (from 5 Li to 11 Li)
Plugging in the volume values:
Work(B to C) = (-0.25/2)(11^2 - 5^2) + 7.25(11 - 5)
Finally, add the work done in process A to B and the work done in process B to C to find the total work done from A to C:
Total Work = Work(A to B) + Work(B to C)