4 mole of di atomic gas is heated at constant pressure from 270 c to 300c change of entropy of the gas is

Change in entropy =∫dQT=nCPIn T2T1=472RIn 600300 =14 R In 2=19.4 cal/K

4 mole of di atomic gas is heated at constant pressure from 270 c to 300c change of entropy of the gas is

Change in entropy =∫dQT=nCPIn T2T1=472RIn 600300 =14 R In 2=19.4 cal/K

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4 mole of di atomic gas is heated at constant pressure from 27⁰ c to 30⁰cchange of entropy of the gas is

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19.4 cal/K

15.2 cal/K

22.8 cal/K

5.6 cal/K

answer is A.

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Change in entropy $= \int \frac{dQ}{T} = {nC}_{P} In \frac{T_{2}}{T_{1}} = 4 (\frac{7}{2} R) In (\frac{600}{300}) = 14 R In 2 = 19.4 cal / K$

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4 mole of di atomic gas is heated at constant pressure from 27⁰ c to 30⁰cchange of entropy of the gas is