Answer to Question #218568 in Molecular Physics | Thermodynamics for vik

Question

enthalpy of the steam at the entry of condenser is 200 kilo joule enthalpy of water at the exit of condenseris 100 kj what will be the work done by condensor

Answer 1

Enthalpy

H=U+PV

dH=dU+PdV

$H_2-H_1=dU+PdV$

Put value

$200-100=dU+PdV$

Isothermal process

dU=0

Put dU value

$$PdV=H_2-H_1=200-100=100KJ$$

$dW=PdV$

Then

dW=PdV=100KJ

Work done by condensor

dW=100kJ