Correct Answer - Option 1 : 1.03
Concept:
Entropy change during the isothermal process is given by
\({s_2} - {s_1} = R\;ln\left( {\frac{{{V_2}}}{{{V_1}}}} \right)\)
Where R is gas-constant, V1 volume before expansion. V2 volume after expansion.
Calculation:
Given:
R =1.486 KJ/Kg, V1 = 1.5 m3 V2 = 3 m3.
\({s_2} - {s_1} = R\;ln\left( {\frac{{{V_2}}}{{{V_1}}}} \right)\)
\({s_2} - {s_1} = 1.486\;ln\left( {\frac{{{3}}}{{{1.5}}}} \right)\)
\({s_2} - {s_1} = 1.03\)