For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using 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. Choose the correct option.
(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) \Delta U=0 in both cases.

Answers (1)

The answer is (iii) and (iv).

Explanation: Now, q = -w, in the case of an isothermal reversible change

Q=-w=nRT\; In\frac{vf}{vi}

\frac{w(600k)}{w(300k)}=\frac{1\times R\times 600k\; In\frac{10}{1}}{1\times R\times 300K\; In\frac{10}{1}}=2

In case of ideal gases, \Delta U = 0 for the expansion, which is isothermal. There is no change in the internal energy as the temperature is constant, and, hence \Delta U = 0.

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