Question #252325
A closed system undergoes a reversible process at a constant pressure process of 3.5 bar and its volume changes from 0.15 m3to 06m3 . 25 kJ of heat is rejected by the system during the process. Determine the change in internal energy of the system.
1
Expert's answer
2021-10-18T11:08:28-0400

Constant Pressure Process

P = 3.5 bar =3.5×105  Pa= 3.5 \times 10^5\; Pa

V1=0.15  m3V2=0.06  m3V_1 = 0.15 \; m^3 \\ V_2 = 0.06 \; m^3

Heat added = 25 kJ = 25000 J

For Constant Pressure Process, Work Done is given as

W=P(V2V1)W=3.5×105(0.060.15)W=31500  JW = P(V_2 – V_1) \\ W = 3.5 \times 10^5 (0.06 - 0.15) \\ W = – 31500 \; J

Negative sign indicates that the work is done on the system

Internal Energy

From First Law we have

ΔU = Q – W

ΔU = 25000 – (– 31500)

ΔU = 25000 + 31500

ΔU = 56500 J

Change in Internal Energy = 56500 J


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