JEE CHEMISTRY MCQ: THERMODYNAMICS

1 Hard

1 of 1

Application +4 | -1

A system undergoes a certain change in state by path I and the corresponding heat absorbed and work done are

10 \mathrm{~kcal}

and

0 \text{ erg},

respectively. For the same change in state by path II, the respective quantities are

11 \mathrm{~kcal}

and

0.5 \, w_{\max },

where

w_{\max }

represents the work done if the changes were reversibly carried out. The magnitude of

w_{\max },

(1 \mathrm{~cal}=4.2 \mathrm{~J})

View answer and explanation

Correct option: A

8400 \mathrm{~J}

Option A (8400 J) is correct because it is derived from equating the change in internal energy for both paths and solving for the maximum work.

More Questions from THERMODYNAMICS

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\ \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}