Answer
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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.
Additional information:
The change in free energy occurs when a compound is formed from its elements in their most thermodynamically stable state at standard state conditions i.e. 1 atm.In thermodynamics, the Gibbs free energy is a thermodynamic potential that can be used to calculate the maximum of irreversible work that may be performed by a thermodynamic system at a constant temperature and pressure.The Gibbs free energy is the maximum amount of non-expansion work that can be extracted from a thermodynamic closed system. This maximum can be attained only in a completely reversible process.
Note: Enthalpy of vaporization is the amount of energy that must be added to the liquid substance, to transform a quantity of that substance into gas. Entropy of vaporization is an increase in entropy upon vaporization of a liquid.
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.
Additional information:
The change in free energy occurs when a compound is formed from its elements in their most thermodynamically stable state at standard state conditions i.e. 1 atm.In thermodynamics, the Gibbs free energy is a thermodynamic potential that can be used to calculate the maximum of irreversible work that may be performed by a thermodynamic system at a constant temperature and pressure.The Gibbs free energy is the maximum amount of non-expansion work that can be extracted from a thermodynamic closed system. This maximum can be attained only in a completely reversible process.
Note: Enthalpy of vaporization is the amount of energy that must be added to the liquid substance, to transform a quantity of that substance into gas. Entropy of vaporization is an increase in entropy upon vaporization of a liquid.
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