A fluid contained in a cylinder receives 150 kj of mechanical energy by means of paddle wheel, together with 50 kj of heat. At the same time, the piston in the cylinder moves at a constant pressure of 200 kpa during the fluid expansion from 2 m3 to 3 m3. Determine the change in the internal energy and the enthalpy of the fluid.
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My name is chaurasia Icon
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Hey mate,
◆ Answer-
∆H = 200 kJ
∆U = -400 kJ
● Explanation-
# Given-
Qw = 150 kJ
Qh = 50 kJ
P = 200 kPa
V1 = 2 m^3
V2 = 5 m^3
# Solution-
Here, total energy input will be-
Q = Qw + Qh
Q = 150 + 50
Q = 200 kJ
Here, change in enthalpy is equal to energy input.
i.e. ∆H = Q = 200 kJ
By 1st law of thermodynamics,
∆H = ∆U + P∆V
∆U = ∆H - P∆V
∆U = 200 - 200(5-3)
∆U = 200 - 600
∆U = -400 kJ
Therefore,
Change in internal energy is -400 kJ and change in enthalpy is 200 kJ.
Hope that is useful...
◆ Answer-
∆H = 200 kJ
∆U = -400 kJ
● Explanation-
# Given-
Qw = 150 kJ
Qh = 50 kJ
P = 200 kPa
V1 = 2 m^3
V2 = 5 m^3
# Solution-
Here, total energy input will be-
Q = Qw + Qh
Q = 150 + 50
Q = 200 kJ
Here, change in enthalpy is equal to energy input.
i.e. ∆H = Q = 200 kJ
By 1st law of thermodynamics,
∆H = ∆U + P∆V
∆U = ∆H - P∆V
∆U = 200 - 200(5-3)
∆U = 200 - 600
∆U = -400 kJ
Therefore,
Change in internal energy is -400 kJ and change in enthalpy is 200 kJ.
Hope that is useful...
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