The degree of irreversibility of a process may be given as q irr / q rev . The entropy change for an isothermal expansion of an ideal gas at 300 K from a particular state A to another state B is 5

Question

The degree of irreversibility of a process may be given as $q_{irr} / q_{rev}$ . The entropy change for an isothermal expansion of an ideal gas at 300 K from a particular state A to another state B is $50 J / K$ . When expansion was performed at constant external pressure, the work done by the system was 6 KJ . Calculate the degree of irreversibility.

Answer:

0

Solution:

Entropy change

$Δ S = 50 J / K$ .

$q_{re v} = T Δ S$ .

$q_{re v} = 300 \times 50 = 15000 J$

work = $6000 J$ .

$\frac{q _{i rr}}{q _{re v}} = \frac{6000}{15000} = 0.40$

TestHub · Accepted Answer

Entropy change Δ S = 50 J / K . q re v ​ = T Δ S . q re v ​ = 300 × 50 = 15000 J work = 6000 J . q re v ​ q i rr ​ ​ = 15000 6000 ​ = 0.40

Chemistry - Thermodynamics - II Question with Solution | TestHub

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