Answer to Question #252325 in Molecular Physics | Thermodynamics for Aravind

Question

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.

Answer 1

Constant Pressure Process

P = 3.5 bar $= 3.5 \times 10^5\; Pa$

$V_1 = 0.15 \; m^3 \\ V_2 = 0.06 \; m^3$

Heat added = 25 kJ = 25000 J

For Constant Pressure Process, Work Done is given as

$W = P(V_2 – V_1) \\ W = 3.5 \times 10^5 (0.06 - 0.15) \\ W = – 31500 \; J$

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