GATE Mechanical Engineering (ME) 2019 Shift 1 Solved Paper

Question : 20 of 65

Marks: +1, -0

Air of mass

1 {kg}

, initially at

300 {K}

and 10 bar, is allowed to expand isothermally till it reaches a pressure of 1 bar. Assuming air as an ideal gas with gas constant of

0.287 {kJ} {/} {kg} . {K}

, the change in entropy of air (in

{kJ} {/} {kg} ⋅ {K}

, round off to two decimal places) is

Solution:

∆ s=C_{p} \ln \;{T_{2}}/{T_{1}}-R \ln \;{P_{2}}/{P_{1}}

As the expansion is isothermal, so

T_{1}=T_{2}

∆ s=-R \ln \;{P_{2}}/{P_{1}}=-0.287 × \ln \;{1}/{10}

=-0.287 ×-2.3026=0.661

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