The latent heat of vaporization of water is 540 cal $\mathrm{g}^{-1}$ at $100^{\circ} \mathrm{C}$. What will be the entropy increase when one mole of water is evaporated at $100^{\circ} \mathrm{C}$ ?

$28 \mathrm{~cal} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

$540 \mathrm{~cal} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

$1.82 \mathrm{~cal} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

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JEE CHEMISTRY MCQ: THERMODYNAMICS

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Application +4 | -1

The latent heat of vaporization of water is 540 cal

\mathrm{g}^{-1}

100^{\circ} \mathrm{C}

. What will be the entropy increase when one mole of water is evaporated at

100^{\circ} \mathrm{C}

View answer and explanation

Correct option: B

26 \mathrm{~cal} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

The correct answer is B (26 cal K^-1 mol^-1) because it is calculated using the formula ΔS = ΔH/T, where ΔH is the molar heat of vaporization and T is the temperature in Kelvin.

More Questions from THERMODYNAMICS

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Consider following reactions
I:

\ \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}(\mathrm{g}), \Delta \mathrm{H}_1=\mathrm{a}

II:

\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}_2(\mathrm{~g}), \Delta \mathrm{H}_2=\mathrm{b}

III:

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})-2 \mathrm{CO}(\mathrm{g}) ; \Delta \mathrm{H}_3=\mathrm{c}

Select the correct statement When

0.18 \mathrm{~g}

of glucose was burnt in a bomb calorimeter, the temperature rose by

4 \mathrm{~K}.

The heat capacity of the calorimeter system is

\left(\Delta_{\mathrm{C}} \text{H}^{\circ}=-2.8\times 10^6 \mathrm{~J} / \mathrm{mol} \text{ glucose}\right)

Statement-1:

{CsI}

is less soluble in water compared to

{CsCl}

Satement-2: More amount of heat is liberated during the formation of

{CsCl}

than the formation

{Cs} I

from their elements Consider the following processes :

\begin{array}{cccccccccccccccccccc} {}&{\Delta H(KJ/mol)} \\ {1/2A \to B}&{ + 150} \\ {3B \to 2C + D}&{ - 125} \\ {E + A \to 2D}&{ + 350} \end{array}

For

{B}+{D} \rightarrow {E}+2 {C}, \Delta {H}

will be : The difference b/w

\Delta H

and

\Delta u

at constant volume is equal to The heat evolved in the combustion of methane is given by the following equations

\mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right) + 2{\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( l \right);\Delta H = - 890.3\mathrm{kJ}

How many grams of methane would be required to produce 44.15 kJ of heat of combustion?