Saved Bookmarks
| 1. |
Write expression for the work done by 1 moleof the gas in each of the following cases : (i) For irreversible expansion of the gas from volume V_(1) to V_(2). (ii) For reversible isothermal expansion of the gas from volume V_(1) toV_(2). (iii)For expansion of the gas into an evaluated vessel. (iv) For reversibleisothermal compression of the gas from pressure P_(1) to P_(2) (v)For adiabatic expansion resulting into changeof temperature from T_(1) to T_(2). |
|
Answer» Solution :(i)IRREVERSIBLE EXPANSION takes place when external pressure `(P_(ext))` remains constant `w_("irrev")= - P_("ext") (V_(2)-V_(1))= -P_(ext) DeltaV` (ii)Reversible expansion takes place when internal pressure is infinitesimally greater than exterhanl pressure `(P_("int")~=P_(ext))` at EVERY stage. Thsu, externalpressurehas to be ADJUSTED throughout. `w_(rev)= -nRT LN. (V_(2))/(V_(1))` (iii) The expansion is irreversible. Further, as `P_(ext) =0`, therefore, `w= -P_(ext)DeltaV= 0 xx ( DeltaV) = 0` (iv) When gas is compressed, work is done on the gas. For isothermalreversible compression, `w= + nRT ln. (P_(2))/(P_(1))` (v) For adiabatic expansion, temperature falls, i.e.,`T_(2) ltT_(1))` `w=C_(v) (T_(2)-T_(1)),i.e., w` is -ve. |
|