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Question
For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using the expression w = `- nRT` In `V_f/V_i`. A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversibly to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively.
(i) Work done at 600 K is 20 times the work done at 300 K.
(ii) Work done at 300 K is twice the work done at 600 K.
(iii) Work done at 600 K is twice the work done at 300 K.
(iv) ∆U = 0 in both cases.
Solution
(iii) Work done at 600 K is twice the work done at 300 K.
(iv) ∆U = 0 in both cases.
Explanation:
For isothermal reversible change, we know
q = – w = `nRT` In `V_f/V_i` = 2.303 `nRT` log `V_f/V_i`
`(W_(600 K))/(W_(300 K)) = (1 xx R xx 600 K In 1/10)/(1 xx R xx 300 K In 1/10) = 600/300` = 2
For isothermal expansion of ideal gases, ΔU = 0.
Since temperature is constant this means there is no change in internal energy.
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