Question

one mole of an ideal gas at 300 K expands isothermally and reversibly from 5 to 20 litres. calculate the work done and heat absorbed by the gas

Answer 1

Here, Temperature, T = 300K
Final volume, V_f = 20 l
Initial volume, V_i = 5 l

Work done in an isothermal reversible process is given by:
$\mathrm{W}=\mathrm{nRT}\ln \frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}$

$$\begin{gathered} =1\mathrm{mol}\times 8.314{\mathrm{Jmol}}^{-1}{\mathrm{K}}^{-1}\times 300\mathrm{K}\times \ln \frac{20}{5} \\ =3457.69\mathrm{J} \\ =3.4577\mathrm{kJ} \end{gathered}$$

Now, From first law of thermodynamics,
$∆\mathrm{U}=\mathrm{Q}+\mathrm{W}$

Since, temperature is constant , $∆\mathrm{U}$ = 0, and hence Q = -W.
Therefore the heat absorbed by the gas is equal to the amount of work done.
This implies that heat absorbed by the gas, Q = -3.4577 kJ.