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JEE CHEMISTRY MCQ: THERMODYNAMICS

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Understanding +4 | -1
Read the following statements carefully and choose the correct option

(i) The state of the system is specified by state functions or state variables.

(ii) Variables like P,V\mathrm{P}, \mathrm{V} and T\mathrm{T} are called state variables or state functions

(iii) Their values depend only on the state of the system and not on how it is reached.
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Correct option: C
(i), (ii) and (iii) are correct
Option C ((i), (ii) and (iii) are correct) is correct because all three statements accurately describe state functions and state variables.

More Questions from THERMODYNAMICS

Practice All Chapter MCQs β†’
Consider following reactions
I: Β C(s)+12O2(Β g)βˆ’CO(g),Ξ”H1=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(Β g)βˆ’CO2(Β g),Ξ”H2=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(Β g)βˆ’2CO(g);Ξ”H3=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 0.18Β g0.18 \mathrm{~g} of glucose was burnt in a bomb calorimeter, the temperature rose by 4Β K.4 \mathrm{~K}. The heat capacity of the calorimeter system is (Ξ”CH∘=βˆ’2.8Γ—106Β J/molΒ 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 :

Ξ”H(KJ/mol)1/2Aβ†’B+1503Bβ†’2C+Dβˆ’125E+Aβ†’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 Ξ”H\Delta H and Ξ”u\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)β†’CO2(g)+H2O(l);Ξ”H=βˆ’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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