Question

0.1 mol of ideal gas compressed reversibly isothermally from 10 l to 1 l at 100 K. The work done in the process is-

Answer 1

Step 1: Formula of work done

The work done in the isothermal reversible compression is given by the formula:

$\mathrm{W}=-\mathrm{nRTln}\left(\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}\right)$

Where W = work done, n = no. of moles, R = gas constant = 8.314, T = temperature , ${\mathrm{V}}_{\mathrm{f}}$ = Final volume of gas, ${\mathrm{V}}_{\mathrm{i}}$ = Initial volume of gas

Step 2: Calculating the work done

Put the values in the formula, i.e.:

n = 0.1, R = 8.314, T = 300K, ${\mathrm{V}}_{\mathrm{f}}$ = 1l, and ${\mathrm{V}}_{\mathrm{i}}$ = 10l

$$\begin{gathered} \mathrm{W}=-0.1\times 8.314\times 300\times \ln \left(\frac{1}{10}\right) \\ =115.23\mathrm{kJ} \end{gathered}$$

Hence, the work done in the process is 115.23 kJ.