Shown in the figure, an insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains an indeal gas, and the other part is evacuated. The partition is then removed, and the gas expands into the entire tank. At the initial state, the mass of the gas is m= 4.00kg, initial pressure is p1 = 600.00 kPa, initial temperature is T1 = 300.00 K. The gas constant is R = 0.7180 kJ/(kg·K). (The internal energy can be determined by the equation ΔU=m·cv·(T2-T1), where cv = 0.7180 kJ/(kg·K) is the specific heat at the constant volume.)Calculate the internal energy change ΔU.__________ (kJ) Shown in the figure, an insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains an indeal gas, and the other part isevacuated. The partition is then removed, and the gas expands into the entire tank. At the initial state, the mass of the gas is m= 4.00kg, initial pressureis p1 = 600.00 kPa, initial temperature is T1 = 300.00 K. The gas constant is R = 0.7180 kJ/(kg-K). (The internal energy can be determined by theequation AU=m-cy (T2-T1), where cy= 0.7180 kJ/(kg-K) is the specific heat at the constant volume.)Calculate the internal energy change AU.Ideal gasT1FISEP1V1mState 1Evacuated(kJ)Ideal gasT2=?P2=?V2=2V1State 2

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Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Shown in the figure, an insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains an indeal gas, and the other part is evacuated. The partition is then removed, and the gas expands into the entire tank. At the initial state, the mass of the gas is m= 4.00kg, initial pressure is p1 = 600.00 kPa, initial temperature is T1 = 300.00 K. The gas constant is R = 0.7180 kJ/(kg·K). (The internal energy can be determined by the equation ΔU=m·c_v·(T₂-T₁), where c_v= 0.7180 kJ/(kg·K)is the specific heat at the constant volume.)

Calculate the internal energy change ΔU.__________ (kJ)

Shown in the figure, an insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains an indeal gas, and the other part is
evacuated. The partition is then removed, and the gas expands into the entire tank. At the initial state, the mass of the gas is m= 4.00kg, initial pressure
is p1 = 600.00 kPa, initial temperature is T1 = 300.00 K. The gas constant is R = 0.7180 kJ/(kg-K). (The internal energy can be determined by the
equation AU=m-cy (T2-T1), where cy= 0.7180 kJ/(kg-K) is the specific heat at the constant volume.)
Calculate the internal energy change AU.
Ideal gas
T1
FISE
P1
V1
m
State 1
Evacuated
(kJ)
Ideal gas
T2=?
P2=?
V2=2V1
State 2

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