Question

one mole of water at 100 degree celsius is converted into steam at 100 degree Celsius at a constant pressure of 1 ATM. The change in entropy is [heat of vaporization of water at 100 degree celsius is 540 calorie per gram]


a) 8.74
b) 18.76
c) 24.06
d) 26.06​

Answer 1

Answer:

d

Here Your Solution

∆S = Q /T

Q = mL

m = 18g (1 mole Given)

L = 540cal/gram

T = 373 Kelvin (BP)

∆S = 18 × 540/273

∆S = 26.06