InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
One mole of a liquid ( 1 bar, 100 ml) is taken in an adiabatic container and the pressure increases steeply to 100 bar. Then at constant pressure of 100 bar, volume decrases by 1 ml.Find `DeltaU` and`DeltaH`. |
|
Answer» As the process is carriedout under adiabatic condition, `q=0` . But `DeltaU = q + w`. Hence `Delta U =w`.Further, at constant pressure of100 bar, volume has decreased by 1 ml, therefore, work of contraction . `= P Delta V = 100 ` bar ` xx 1 `ml. `= ( 100 xx10^(5) Nm^(-2)) ( 10^(-6)m^(3)) = 10J` Hence, `Delta U = 10 J` `Delta H =Delta U + Delta( PV)` `Delta ( PV) = (P_92)V_(2) - P_(1)V_(1))` But `P_(1) =1 "bar", V+(1) = 100 ml` `P_(2) = 100 "bar", V_(2) = 99 ml` `:. Delta ( PV) = ( 100 "bar" xx 99 ml) - ( 1 "bar" xx 100 ml) `= ( 9900 - 100 ) "bar" ml = 9800 "bar ml ) = 980 J ` `:. Delta H = 10J + 980 J = 990 J` |
|