An insulated rigid tank contains 1.5 kg of helium at 30∘C and 500 kPa. A paddle wheel with a power rating of 0.1 kW is operated within the tank for 30 minutes. Cvof helium = 3.12 KJ/Kg.K
Problem 7A. Calculate the net work done by helium, in kJ. Enter the number with appropriate sign.
Problem 7B. What is the final temperature of helium, in ∘C ?
Problem 7C. What is the final pressure of helium, in kPa?
Problem 7D. What is the change in entropy of helium, in J/K?
Problem 7E. How much entropy is produced during the process, in J/K?
Problem 7F. What are the causes of the entropy production?
a)Violation of the second law
b)An adiabatic process
c)The stirrer work
d)The increase in temperature
-180
69 C
563.5
550
550
b)
all wrong
all correct except last one. option is wrong/more than one option should be included.
To calculate the different values in this problem, we need to apply the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
ΔU = Q - W (Equation 1)
First, let's calculate the work done by the helium (Problem 7A):
Given information:
Mass of helium (m) = 1.5 kg
Initial temperature (T1) = 30 ∘C = 30 + 273.15 K
Initial pressure (P1) = 500 kPa
Power rating of the paddle wheel (P) = 0.1 kW
Time (t) = 30 minutes = 30 × 60 s
We know that work done (W) is equal to the power (P) multiplied by the time (t):
W = P × t (Equation 2)
Let's calculate W:
W = 0.1 kW × 30 × 60 s = 180 kJ
We need to convert this to kJ since the heat and internal energy changes are usually measured in kJ.
Now, let's calculate the change in internal energy (ΔU) using Equation 1.
The change in internal energy is given by:
ΔU = m × Cv × ΔT (Equation 3)
where m is the mass of helium, Cv is the specific heat at constant volume for helium, and ΔT is the change in temperature.
Given information:
Cv of helium (Cv) = 3.12 kJ/kg.K
To find ΔT, we need to use the ideal gas law:
P1V1/T1 = P2V2/T2 (Equation 4)
Since the tank is rigid and insulated, the volume (V) remains constant. Hence, equation 4 simplifies to:
P1/T1 = P2/T2 (Equation 5)
Rearranging Equation 5, we can find the final temperature (T2) of the helium in terms of the given values:
T2 = T1 × (P2/P1)
Let's calculate T2 using Equation 5:
T2 = (30 + 273.15) K × (500 kPa/P1)
Let's calculate T2 using the given values and then calculate the change in internal energy (ΔU) using Equation 3.
Once we have the value of ΔU, we can now calculate the heat added to the system (Q) using Equation 1:
Q = ΔU + W
Let's calculate Q.
Problem 7A: The net work done by helium is 180 kJ (positive value since work is done by the gas).
Problem 7B: The final temperature of helium can be obtained by solving Equation 5 for T2 using the given values and then converting it to ∘C.
Problem 7C: The final pressure of helium remains constant at 500 kPa.
Problem 7D: The change in entropy (ΔS) can be calculated as ΔS = Q/T using the heat added to the system (Q) and the final temperature (T2). Express the result in J/K.
Problem 7E: The entropy produced during the process can be calculated by subtracting the initial entropy from the final entropy. The initial entropy is 0 since the helium is initially at rest. Express the result in J/K.
Problem 7F: The causes of entropy production can be attributed to the stirring work (the paddle wheel doing work on the system) and the increase in temperature.
210
70
12
1000
800
the increase in temperature