11th Class Chemistry Thermodynamics

  • question_answer 53)   For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using the expression \[w=-nRT\ln \frac{{{V}_{f}}}{{{V}_{i}}}.\] 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. 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:

      (c, d) Work done in isothermal process can be calculated as: \[w=2.303nRT\,\log \left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\] \[\frac{{{w}_{600}}}{{{w}_{300}}}=\frac{2.303\times 1\times R\times 600\log (10)}{2.303\times 1\times R\times 300\log (10)}=2\] \[\Delta U=0\]for isothermal processes.                


You need to login to perform this action.
You will be redirected in 3 sec spinner