InterviewSolution
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InterviewSolution
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For an ideal gas, 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.A. Work done at 600 K is 20 times at the work done at 300KB. Work done at 300 K is twice the work done at 600 KC. Work done at 600 K is twice the work done at 300 KD. `DeltaU=0`in both cases |
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Answer» Correct Answer - c,d `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` |
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