Answer:
-669.2 J/k
Explanation:
We are given that
Mass=m=0.135 kg
Normal boiling point,B.P=4.216 K
Heat f of vaporization=[tex]H_v=-2.09\times 10^4 J/kg[/tex]
We have to find the change in entropy of helium gas.
[tex]\Delta S=\frac{mH_v}{T}[/tex]
Using the formula then, we get
[tex]\Delta S=\frac{0.135\times (-2.09\times 10^4}{4.216}[/tex]
[tex]\Delta S=-669.2 J/K[/tex]
Hence, the change in entropy of helium gas=-669.2 J/k
