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JEE CHEMISTRY MCQ: THERMODYNAMICS

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Application +4 | -1

Geological conditions are sometimes so extreme that quantities neglected in normal laboratory experiments take on an overriding importance. For example, consider the formation of diamond under geophysically typical conditions. The density of graphite is

2.4 \mathrm{~g} / \mathrm{cm}^3

and that of diamond is

3.6 \, \mathrm{g} / \mathrm{cm}^3

at a certain temperature and

500 \mathrm{~kbar}.

By how much does

\Delta U_{\text {trans}}

differs from

\Delta H_{\text {trans}}

for the graphite to diamond transition?

View answer and explanation

Correct option: A

83.33 \mathrm{~kJ} / \mathrm{mol}

Option A (83.33 kJ/mol) is correct because it calculates the difference between \

\Delta H

and \

\Delta U

using the given pressure and volume change during the graphite to diamond transition.

More Questions from THERMODYNAMICS

Practice All Chapter MCQs →

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}

\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}

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)

{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}

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

will be : The difference b/w

\Delta H

\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?