Discussion Forum

One mole of an ideal gas at 300 k  is expanded isothermally from an initial volume of 1 litre to 10 litres. Then $\triangle$ S  (cal deg^-1  mol^-1) for this process is : (R=2 cal K^-1 mol^-1)

Answer:

Explanation:

For an isothermal process

$\triangle S= 2.303 nR \log\frac{V_{2}}{V_{1}}$

  $\triangle S= 2.303 \times1\times2\times\log \frac{10}{1}$

 = 4.6 cal deg^-1 mol^-1