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JEE CHEMISTRY MCQ: THERMODYNAMICS

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Understanding +4 | -1
Assertion :- The value of specific heat at constant pressure is more than that at constant
volume.

Reason :- The energy required to raise temperature by a unit at constant pressure is
greater because some amount of heat is used in doing work.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A is correct because the specific heat at constant pressure is higher than at constant volume due to the additional energy required for the system to do work against the external pressure.

More Questions from THERMODYNAMICS

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Consider following reactions
I:  C(s)+12O2( g)−CO(g),ΔH1=a\ \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}(\mathrm{g}), \Delta \mathrm{H}_1=\mathrm{a}

II: CO(g)+12O2( g)−CO2( g),ΔH2=b\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}_2(\mathrm{~g}), \Delta \mathrm{H}_2=\mathrm{b}

III: C(s)+CO2( g)−2CO(g);ΔH3=c\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 g0.18 \mathrm{~g} of glucose was burnt in a bomb calorimeter, the temperature rose by 4 K.4 \mathrm{~K}. The heat capacity of the calorimeter system is (ΔCH∘=−2.8×106 J/mol glucose)\left(\Delta_{\mathrm{C}} \text{H}^{\circ}=-2.8\times 10^6 \mathrm{~J} / \mathrm{mol} \text{ glucose}\right) Statement-1: CsI{CsI} is less soluble in water compared to CsCl{CsCl}
Satement-2: More amount of heat is liberated during the formation of CsCl{CsCl} than the formation CsI{Cs} I from their elements
Consider the following processes :

ΔH(KJ/mol)1/2A→B+1503B→2C+D−125E+A→2D+350\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→E+2C,ΔH{B}+{D} \rightarrow {E}+2 {C}, \Delta {H} will be :
The difference b/w ΔH\Delta H and Δu\Delta u at constant volume is equal to The heat evolved in the combustion of methane is given by the following equations CH4(g)+2O2(g)→CO2(g)+H2O(l);ΔH=−890.3kJ\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?
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