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  •  2 moles of an ideal gas <img alt="\left ( C_{v,m}=15JK^{-1}mol^{-1} \right )" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%20%28%20C_%7Bv%2Cm%7D%3D15JK%5E%7B-1%7Dmol

 2 moles of an ideal gas \left ( C_{v,m}=15JK^{-1}mol^{-1} \right )  are at 300 \:K and 5 \:L. If the gas is heated to 350\: Kand the volume changed to 15 \:L , Calculate the entropy change.

Option: 1

26.76  J/K


Option: 2

4.62 J/K


Option: 3

31.38 J/K


Option: 4

22.14 J/K


Answers (1)

best_answer

We have given,

Initial volume,V_{1} = 5L

Final volume, V_{2} = 15L

Initial Temperature, T_{1} = 300\: K

Final Temperature, T_{2} = 350 \:K

Number of moles, n = 2

Change in entropy as a function of T and V is given as, 

\Delta S=nC_{V.m}\:In\:\frac{T_{2}}{T_{1}}+nR\:In\:\frac{V_{2}}{V_{1}}

\Delta S=2\times15\times\:In\:\frac{T_{2}}{T_{1}}+2\times8.314\times\:In\:\frac{V_{2}}{V_{1}}

\Delta S=4.62+26.76

\Delta S=31.38\:J/K

Hence, option number ( 3 ) is correct 

Posted by

Deependra Verma

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