For a particular gaseous system (not necessarily an ideal gas!) it has been dete

Question

For a particular gaseous system (not necessarily an ideal gas!) it has been determined that the energy is given by

U=1.5PV+const.

The system is initially in the state

P=1MPa, V=0.01 m^3;

Designated as point A in the figure.

The system is taken through the cycle of three processes

(A--?B, B--?C, and C--?A)

Shown in the figure.

1. Calculate Q and W for each of the three processes.

2. For the cyclic process A--?B--?C--?A, calculate the total work done on the gas and the total amount of heat received by the gas. What is the change of the internal energy for a complete cycle?

3. Calculate Q and W for a process from A to B along the parabola

P=5*10^5+5*10^9*|V-0.02|^2,where all quantities are given in SI units.

For a particular gaseous system (not necessarily an ideal gas!) it has been determined that the energy is given by U=1.5PV+const. The system is initially in the state P=1MPa, V=0.01 m^3; Designated as point A in the figure. The system is taken through the cycle of three processes (A--?B, B--?C, and C--?A) Shown in the figure. Calculate Q and W for each of the three processes. For the cyclic process A--?B--?C--?A, calculate the total work done on the gas and the total amount of heat received by the gas. What is the change of the internal energy for a complete cycle? Calculate Q and W for a process from A to B along the parabola P=5*10^5+5*10^9*|V-0.02|^2,where all quantities are given in SI units.

Explanation / Answer

for the overall process:

work = area inside the cycle = 3*0.02/2 * 10^6 joules= 3 * 10^4 Joules

and Q=-W (since internal energy U is constant for the whole cycle)

=> Q = -3* 10^4 joules

for the process A-B

work = area under AB line = 0.02*1 *10^6 joules = 2 * 10^4 Joules

U(B)-U(A) = Q + W

P(B)V(B) + const - P(A)V(A) - const = Q + 2 * 10^4

=> (0.03 * 10^6 - 0.01*10^6) = Q + 2 * 10^4

=> Q = 0 Joule

for process B-C

work = area under BC line = 3*0.02/2 * 10^6 + 0.02*1 *10^6 joules = 5 * 10^4 Joules

U(C)-U(B) = Q + W

P(C)V(C) + const - P(B)V(B) - const = Q + 5* 10^4

=> (0.03 * 10^6 - 4*0.01*10^6) = Q + 5 * 10^4

=> -0.01*10^6 = Q + 5 *10^4

=> Q = -6 * 10^4 joules

for process C-A

work = area under BC line = 0 Joules

U(A)-U(C) = Q + W

P(A)V(A) + const - P(C)V(C) - const = Q

=> 0.01*10^6 - 0.03 * 10^6 = Q

=> Q = -2*10^4

work =

area under the parabola = integration from V(A) to V(B) of { 5*10^5+5*10^9*|V-0.02|^2 }

=> { 5*10^5*V(B) + (5*10^9*|V(B)-0.02|^3 )/3 } - { 5*10^5*V(A) + (5*10^9*|V(A)-0.02|^3 )/3 }

=> { 5*10^5*0.03 + (5*10^9*|0.03-0.02|^3 )/3 } - { 5*10^5*0.01 + (5*10^9*|0.01-0.02|^3 )/3 }

=> 15*10^3 - 5*10^3/3 - 5*10^3 - (-5*10^3)/3

=> 10^2 - 10^2/3

=> 66.66 Joules

U(B)-U(A) = Q + W

P(B)V(B) + const - P(A)V(A) - const = Q + 66.66

=> (0.03 * 10^6 - 0.01*10^6) = Q + 66.66 joules

=> Q = 20000 - 66.66 joules = 19933.34 joules

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