For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using expression ` = -nRT ln.(V_(f))/(V_(i))`. A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversible to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option

Options

A. Work done at 600 K is 20 times at the work done at 300K
B. Work done at 300 K is twice the work done at 600 K
C. Work done at 600 K is twice the work done at 300 K
D. `DeltaU=0`in both cases

Answer

c,d

Explanation

`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 the work done at 300K.
As each case involves isothermal expansion of an ideal gas, there is no change in internal energy, i.e.,
`DeltaU =0`