A sample containing 2.000 mol of helium gas originalyy at 298.15 K and 1.000 bar is coled to its normal boilding temperature of 4K, condensed to a liquid and then cooled further to 2K, where it undergoes another phase transition to a second liquid form, called liquid helium II. A beam of laser light suddenly helium II. A beam of laser light suddenly vaporizes this liquid phase, and the helium is brought to a temperature of 298.15K and a pressure of 0.500 bar. the entropy change for the above process is

Options

A. `11.5 JK^(-1)`
B. `-11.5 JK^(-1)`
C. `23 JK^(-1)`
D. Insufficient information

Answer

A

Explanation

Entropy, a state function, does not depends on path, Hence, for
`He(g) [2 mol, 398.15 K, 1.00 " bar" ]rarr rarr rarr`
He (g) [2 mol, 295 K, 0.50 bar]
`rArr Delta S = nR ln. (P_(1))/(P_(2)) 2 R ln 2 = 11.5 JK^(-1)`