Answers)
i)
ΔS= nCv ln(T2/T1) + nR ln(V2/V1)
ΔS= n Cv ln (40+273.15/-15+273.15) + nR ln ( 9/5)
ΔS= n Cv *0.19 + nR *0.58
As both the quantities are positive, ΔS >0
ii)
ΔS= nCv ln(T2/T1) + nR ln(V2/V1)
ΔS= n Cv ln (-18+273.15/-18+273.15) + nR ln ( 8/13)
ΔS= 0 + nR *(-0.48)
ΔS= - 0.48 nR
ΔS<0
iii)
For the given problem, no information is provided for the final Volume of water.
So, no enough information.
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