a fluid undergoes a reversible adiabatic compression from .5 MPa, .2 m3 to .05 m3 according to the law PV1.3 = constant. determine the change in enthalpy, internal energy and work transfer during the process.
P1V11.3 = P2V21.3
dH = T dS + V dP
dS = 0, due to reversible - adiabatic process.
dH = ∫V dP
dU = dH - (P2V2 - P1V1)
W = dU
The Attempt at a Solution
when i solve the integral equation of enthalpy, i found its value H2-H1 = 223.3 kJ
U2-U1 = 171.77 kJ.
i didn't understand the conceptual part of the problem.
gas is being compressed by doing work on the system. As it is being compressed, the work done on the system is completely transformed into it internal energy (increase in energy stored by the system).
enthalpy rise = (internal energy rise) + Δ(PV)
= 171.77 kJ + 51.53 kJ
i understood the part that enthalpy of gas is increased due to the rise in internal energy and also the rise internal energy is due to the work done on the system.
but i didn't understand the part, how enthalpy value got increased due to the PV contribution? i mean there is no other energy transfer to the system apart from work transfer and all the energy due to work transfer has gone into rising it's internal energy.
doesn't change in internal energy, enthalpy requires energy transfer (heat ot work) in case of closed system?
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