For an idealgas, the work of reversible expansion under isothermal condition can be calculated by using the expression w= - nRT ln . (V_(f))/( V_(i)0 . A sample containig 1.0 molof an ideal gas is expanded isothermally and reversibly to ten times of its original volume in two separate experiments . The expansions is carried out at 300 K and at600 K respectively. Choose the correct option.

For an idealgas, the work of reversible expansion under isothermal con

1.	For an idealgas, the work of reversible expansion under isothermal condition can be calculated by using the expression w= - nRT ln . (V_(f))/( V_(i)0 . A sample containig 1.0 molof an ideal gas is expanded isothermally and reversibly to ten times of its original volume in two separate experiments . The expansions is carried out at 300 K and at600 K respectively. Choose the correct option.
Answer» WORK done at 600 K is 20 times at the work done at 300K Work done at 300 K is twice the work done at 600 K Work done at 600 K is twicethe work done at 300 K `DeltaU=0`in both cases Solution :`w= -nRT LN. (V_(1))/( V_(2)). `The factors n , R and `ln. (V_(f))/( V_(i))` are same in both the cases. Hence, `(w_(2))/(w_(1)) = (T_(2))/(T_(1)), i.e., (.^(w)600K)/(.^(w)300K)= ( 600K)/( 300K) =2 , i.e., ` work done at 600K is twice thework done at 300K. As each case involves ISOTHERMAL expansion of an ideal gas, there is no change in internal energy, i.e., `DeltaU =0`