JEE CHEMISTRY MCQ: THERMODYNAMICS

1 Hard

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Application +4 | -1

Molar heat capacity of

\mathrm{CD}_2 \mathrm{O}

(deuterated form of formaldehyde) vapour at constant pressure is vapour

14 \mathrm{~cal} / \mathrm{K}-\mathrm{mol}.

The entropy change associated with the cooling of

3.2 \mathrm{~g}

\mathrm{CD}_2 \mathrm{O}

vapour from

1000 \mathrm{~K}

900 \mathrm{~K}

at constant pressure is (assume ideal gas behaviour for

\mathrm{CD}_2 \mathrm{O})

[ln

0.9=-0.1]

View answer and explanation

Correct option: B

-0.14 \mathrm{~cal} / \mathrm{K}

Option B (

-0.14 \mathrm{~cal} / \mathrm{K}

) is correct because the entropy change is calculated using the formula

\Delta S=n \cdot C_{P, m} \cdot \ln \frac{T_2}{T_1}

, and since the temperature decreases, the entropy change is negative.

More Questions from THERMODYNAMICS

Practice All Chapter MCQs →

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

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

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

0.18 \mathrm{~g}

4 \mathrm{~K}.

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

{CsI}

{CsCl}

{CsCl}

{Cs} I

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

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

\Delta H

\Delta u

\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}