Question

Work is done by a substance in reversible nonflow manner during which 32 KJ are added, the pressure remains constant at 200KPa, the volume changes from .5m3 to.25m3. Compute the change in internal energy in KJ.

Answer 1

Given :

The work done(dW)= p(V₂-V1)

Heat added (dQ)=32kJ

Constant pressure (p)= 200kPa

Change in volume(V1-V2)=0.5-0.25=0.25 m³

Change in internal energy(U1-U2)=?

Solution:

The work done during reversible Constant pressure process =(dW)= p(V₂-V1)=200×10³(0.25)=50 kJ

By first law of thermodynamics,

dQ=dU+dW

Change in internal energy(dU)=dQ-dW=32-50=-18kJ (Answer)

Hence , there is reduction in internal 3nergy by 18 kJ