Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The enthalpy of vaporization of a liquid is 30\$KJmo{l^{ - 1}}\$ and the entropy of vaporization is 75\$Jmo{l^{ - 1}}{K^{ - 1}}\$ . Find the boiling point of the liquid at 1 atm.A. 250k B. 400k C. 450k D. 600k

Last updated date: 20th Apr 2024
Total views: 35.4k
Views today: 1.35k
Verified
35.4k+ views
Hint: the fact that at boiling point of the liquid, liquid & its vapor are in equilibrium i.e. vapor is in equilibrium at one atmospheric pressure. We know that at equilibrium, Gibbs energy change is zero. ie \$\Delta G = 0\$.

Complete step by step solution:
It is given that Enthalpy of vaporization is 30 Kilojoules per mole.
i.e. \$\Delta H\$ = enthalpy of vaporization = 30000 \$Jmo{l^{ - 1}}\$
It is also given that Entropy of vaporization is 75\$Jmo{l^{ - 1}}{K^{ - 1}}\$
i.e. \$\Delta S\$ = entropy of vaporization = 75 \$Jmo{l^{ - 1}}{K^{ - 1}}\$
We know at boiling point of the liquid, vapor is in equilibrium (at one atmospheric pressure) which implies that,
Gibbs energy change is equal to zero.
i.e \$\Delta G = 0\$
We know that,
\$\Delta G = \Delta H - T\Delta S\$
Substituting the values in the equation we get,
0 = 30000 – 75T
\$\therefore T = \dfrac{{3000Jmo{l^{ - 1}}}}{{75Jmo{l^{ - 1}}{K^{ - 1}}}}\$

Therefore, we get;
T = 400 K
∴ Boiling point of the liquid at one atmospheric pressure is 400k.
∴ Correct option - B. 400k.