JEE Advanced 2014 · Paper 2 · Chemistry Q29

ChemistrySingle correct+3 / -1Question 29 of 60

For the process
$\ce{H2O(l) -> H2O(g)}$
at $T=100\ ^\circ$ C and 1 atmosphere pressure, the correct choice is

\Delta S_{system}>0

and

\Delta S_{surroundings}>0

\Delta S_{system}>0

and

\Delta S_{surroundings}<0

\Delta S_{system}<0

and

\Delta S_{surroundings}>0

\Delta S_{system}<0

and

\Delta S_{surroundings}<0

Answer and worked solution

Official answer: B

At the normal boiling point liquid and vapour are in equilibrium, so $\Delta S_\text{total}=0$ .

System. Vapour is far more disordered than liquid: $\Delta S_\text{sys}=\dfrac{\Delta H_\text{vap}}{T}>0$ .
Surroundings. They supply the heat of vaporisation: $\Delta S_\text{surr}=-\dfrac{\Delta H_\text{vap}}{T}<0$ .

(B)

Topics:Entropy Gibbs energy and spontaneity

JEE Advanced 2014 · Paper 2, Chemistry question 29