Question

1 mole of an ideal gas at 37°C temperature undergoes expansion isothermally and reversibly from 5 litre to 20 litre. The entropy change in the
process is
11.5 J/mol-K
7.5 J/mol-K
3.5 J/mol-K
6.5 J/mol-K​

Answer 1

Answer:

3.5 j/mol-k hai iska answer

Answer 2

Given info : 1 mole of an ideal gas at 37°C temperature undergoes expansion isothermally and reversibly from 5 litre to 20 litre.

To find : The entropy change in the process is ..

solution : for isothermally expansion,

dQ = dW = ∫PdV

= ∫nRT/V dV [ as P = nRT/V]

= nRT ∫dV/V

= nRT ln[V₂/V₁]

now entropy change, ∆S = dQ/T

= nRT ln[V₂/V₁]/T

= nR ln[V₂/V₁]

here n = 1 mole , R = 8.314 J/mol/K , V₂ = 20 litre and V₁ = 5 litre

so, ∆S = 1 × 8.314 × ln(20/5)

= 8.314 × ln(4)

= 11.5 J/K

Therefore the entropy change in the process is 11.5 J/K.