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A diatomic gas is enclosed in a piston-cylinder arrangement at a pressure of 3 bar and 300 K. The cylinder has a volume of 1.5 m3. The process undergoes isothermal expansion to 3 m3. If the gas has a gas-constant as 1.486 KJ/Kg. using, In2 = 0.693, entropy change during the above process in kJ/kg is:


1. 1.03
2. 1.20
3. 1.65
4. 1.15

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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 mV2 = 3 m3.

\({s_2} - {s_1} = R\;ln\left( {\frac{{{V_2}}}{{{V_1}}}} \right)\)

\({s_2} - {s_1} = 1.486\;ln\left( {\frac{{{3}}}{{{1.5}}}} \right)\)

\({s_2} - {s_1} = 1.03\)

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