Question

The amount work done by 2 moles of an ideal gas at 298 K in reversible isothermal expansion from 10 litres to 20 litres is.

Answer 1

Here, Temperature, T = 298K
Final volume, Vf = 20 l
Initial volume, Vi = 5 l

Work done in an isothermal reversible process is given by:
W = nRT lnVfVi

= 1 mol ×8.314 Jmol−1K−1×298 K ×ln205 = 3457.69 J =3.4577 kJ

Now, From first law of thermodynamics,
∆U = Q+W

Since, temperature is constant , ∆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 = -3.435 kJ.
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