Question

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
A. work done at 600 K is 20 times the work done at 300 K
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. `Delta U = 0` in both cases.

Answer 1

Correct Answer - C::D
Given that, the work of reversible expansion under isothernal condition can be calculated by using the expression
`W = - nRT ln. (V_(1))/(V_(1))`
`V_(1) = 10 V_(i)`
`T_(2) = 600 K`
`T_(1) = 300 K`
Putting these values in above expression
`W_(600 K) = 1 xx R xx 600 ln. (10)/(1)`
`W_(300 K) = 1 xx R xx 300 K ln. (10)/(1)`
Ration `= (W_(600 K))/(W_(300 K)) = (1 xx R xx 600 K ln.(10)/(1))/(1 xx R xx 300 K ln. (10)/(1)) = (600)/(300) = 2`
For isothermal expansion of ideal gases, `Delta U = 0`. Since, temperature is contant this means there is no change in internal energy. Therefore, `Delta U = 0`