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NEET • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Mixed difficulty

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Showing 241-270 of 560 questions

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241 Easy
241 of 560
Remembering +4 | -1
Heat of formation of compound is defined as :
View answer and explanation
Correct option: A
Heat evolved to form one mole of the compound from its elements
Option A is correct because the heat of formation is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states.
242 Easy
242 of 560
Remembering +4 | -1
Mathematical from of 1st Law
View answer and explanation
Correct option: C
Δu=Q+W\Delta u = Q + W
Option C (Δu=Q+W\Delta u = Q + W) is correct because it accurately represents the mathematical formulation of the first law of thermodynamics.
243 Easy
243 of 560
Analyzing +4 | -1
Which of the following gas will liberate maximum heat on combustion per gram?
View answer and explanation
Correct option: A
Methane
Option A (Methane) is correct because it has the highest hydrogen-to-carbon ratio, leading to the greatest heat release per gram upon combustion.
244 Medium
244 of 560
Understanding +4 | -1
In which of the following condition a chemical reaction can not occur?
View answer and explanation
Correct option: C
∆H increase and ∆S decreases
Option C (∆H increase and ∆S decreases) is correct because it leads to a positive ∆G at all temperatures, making the reaction non-spontaneous.
245 Hard
245 of 560
Understanding +4 | -1
Amongst the following statement's regarding adsorption, those that are valid are –
(A) Activation energy is high for chemisorption process
(B) Physisorption converts into chemisorption at high temperature
(C) Chemisorption converts into physisorption at low temperature
(D) With increase in temperature, the equilibrium concentration of adsorbate increases
View answer and explanation
Correct option: B
A, B
Options A and B are valid. Activation energy is generally high for chemisorption, and physisorption can convert to chemisorption at higher temperatures.
246 Medium
246 of 560
Understanding +4 | -1
A positive change in enthalpy occurs in :
View answer and explanation
Correct option: C
MgCO3(s)→MgO(s)+CO2(g)\mathrm{MgC}{\mathrm{O}_3}\left( s \right) \to \mathrm{MgO}\left( s \right) + \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right)
Option C (MgCO3(s)→MgO(s)+CO2(g)\mathrm{MgC}{\mathrm{O}_3}\left( s \right) \to \mathrm{MgO}\left( s \right) + \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right)) is correct because the decomposition of MgCO3 is an endothermic process, requiring energy input and resulting in a positive enthalpy change.
247 Easy
247 of 560
Understanding +4 | -1
In which case, a spontaneous reaction is possible at any temperature?
View answer and explanation
Correct option: A
∆H-ve,∆S+ve
Option A (∆H-ve,∆S+ve) is correct because a negative enthalpy change (exothermic) and a positive entropy change always result in a negative Gibbs free energy change, making the reaction spontaneous at all temperatures.
248 Easy
248 of 560
Understanding +4 | -1
For an ideal gas, the relation between the enthalpy change and internal energy change at constant temperature is given by :
View answer and explanation
Correct option: B
∆H= ∆U+ ∆nRT
Option B (∆H= ∆U+ ∆nRT) is correct because it represents the relationship between enthalpy change and internal energy change for an ideal gas, where ∆n is the change in the number of moles of gas.
249 Easy
249 of 560
Application +4 | -1
Equal volume of C2H2{C_2}{H_2} and H2{H_2} are combusted under identical condition. The ratio of their heat of combustion is:
H2(g)+12O2(g)→H2O(g);ΔH=−241.8   kJ{H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(g);\Delta H = - 241.8\,\,\,kJ
C2H2(g)+212O2(g)→2CO2(g)+H2O(g);  ΔH=−1300   kJ{C_2}{H_2}(g) + 2\frac{1}{2}{O_2}(g) \to 2C{O_2}(g) + {H_2}O(g);\,\,\Delta H = - 1300\,\,\,kJ
View answer and explanation
Correct option: A
5.37/1
Option A (5.37/1) is correct because the ratio of the heat of combustion is calculated by dividing the heat of combustion of C2H2 by the heat of combustion of H2, considering equal volumes imply equal moles.
250 Easy
250 of 560
Application +4 | -1
The change in internal energy, if 60 J60 {~J} of heat is given to the system at constant pressure and 30 J30 {~J} of work is done by the system, will be:
View answer and explanation
Correct option: D
+30 J+30 {~J}
Option D (+30 J) is correct because the change in internal energy is the difference between the heat added to the system and the work done by the system.
251 Easy
251 of 560
Understanding +4 | -1
Equilibrium constant of a reaction is related to:
View answer and explanation
Correct option: A
Standard Gibbs energy change ∆G°
Option A (Standard Gibbs energy change ∆G°) is correct because the equilibrium constant is directly related to the standard Gibbs free energy change by the equation \∆G° = -RTlnK∆G° = -RTlnK.
252 Medium
252 of 560
Remembering +4 | -1
In the exothermic reaction, the enthalpy of
reaction is always:
View answer and explanation
Correct option: C
negative
Option C (negative) is correct because exothermic reactions release heat, resulting in a negative enthalpy change.
253 Easy
253 of 560
Application +4 | -1
For a gas (R/Cv)=0.67\left( {R/{C_v}} \right) = 0.67 the gas is made up of molecules which are:
View answer and explanation
Correct option: A
Monoatomic
Option A (Monoatomic) is correct because for a monoatomic gas, the ratio R/Cv = 2/3 = 0.67.
254 Medium
254 of 560
Remembering +4 | -1
Calorific value of carbohydrates is approximately:
View answer and explanation
Correct option: A
4.0kcal/g4.0\mathrm{ }kcal/g
Option A (4.0 kcal/g) is correct because it is the approximate calorific value of carbohydrates.
255 Medium
255 of 560
Understanding +4 | -1
In an adiabatic process, no transfer of heat takes place between system and surroundings. Choose the corerct option for free expansion of an ideal gas under adiabatic condition from the following.
View answer and explanation
Correct option: C
q=0,  ΔT=0,  w=0q = 0,\,\,\Delta T = 0,\,\,w = 0
Option C (q=0, ΔT=0, w=0q = 0,\,\Delta T = 0,\,w = 0) is correct because in an adiabatic free expansion of an ideal gas, no heat is exchanged (q=0), no work is done (w=0), and the temperature remains constant (ΔT=0).
256 Medium
256 of 560
Application +4 | -1
A gas is allowed to expand in an insulated container against a constant external pressure of 5 atm from an initial volume of 1.75 L to a final volume of 6.75 L. The change in internal energy (ΔU)(\Delta U) of the gas will be
View answer and explanation
Correct option: A
−2.5kJ- 2.5\mathrm{ }kJ
Option A (-2.5 kJ) is correct because the change in internal energy is equal to the heat exchanged, which is the negative of the work done against the constant external pressure in an adiabatic process.
257 Hard
257 of 560
Analyzing +4 | -1
A solution of 500 ml500 \mathrm{~ml} of 2 M−KOH2 \mathrm{~M}-{KOH} is added to 500 ml500 \mathrm{~ml} of 2 M−HCl2 \mathrm{~M}-{HCl} and the mixture is well shaken. The rise in temperature T1T_1 is noted. The experiment is again performed using 250 ml250 \mathrm{~ml} of KOH{KOH} solution and 500 ml500 \mathrm{~ml} of HCl{HCl} solution and rise in temperature T2T_2 is again noted. Which of the following is correct?
View answer and explanation
Correct option: D
T1T_1 is 1.5 times as larger as T2T_2
Option D (T1T_1 is 1.5 times as larger as T2T_2) is correct because the amount of heat released in the first experiment is 1.5 times that of the second experiment, leading to a temperature rise that is 1.5 times greater.
258 Medium
258 of 560
Understanding +4 | -1
Which correctly represents the physical significance of Gibbs energy change?
View answer and explanation
Correct option: D
−ΔG=Wexpansion- \Delta G = {W_{\mathrm{expansion}}}
Option D (−ΔG=Wexpansion- \Delta G = {W_{\mathrm{expansion}}}) is correct because the decrease in Gibbs free energy (-∆G) represents the maximum work that can be obtained from a system during expansion at constant temperature and pressure.
259 Easy
259 of 560
Analyzing +4 | -1
Statement I Due to atomisation of gaseous molecule,
entropy increases.
Statement II Atoms are more ordered than the molecules.
View answer and explanation
Correct option: C
Statement I is correct but Statement II is incorrect
Statement I is correct, but Statement II is incorrect. Atomization increases entropy because atoms have more freedom than molecules.
260 Hard
260 of 560
Analyzing +4 | -1
Work done by a sample of an ideal gas in a process A{A} is double the work done in another process B.{B}. The temperature rises through the same amount in the two processes. If CAC_{{A}} and CBC_{{B}} are the molar heat capacities for the two processes, then
View answer and explanation
Correct option: B
CA>CBC_{{A}}>C_{{B}}
Option B (CA > CB) is correct because work done is related to heat capacity and temperature change. Since work in A is double that in B for the same temperature rise, CA must be greater than CB.
261 Easy
261 of 560
Remembering +4 | -1
In which state matter have highest entropy?
View answer and explanation
Correct option: C
Gas
Option C is correct. Gases have the highest entropy because their molecules have the greatest freedom of movement and occupy the largest volume, leading to the highest degree of disorder.
262 Medium
262 of 560
Remembering +4 | -1
The energy that opposes dissolution of a solvent
is
View answer and explanation
Correct option: B
lattice energy
Option B (lattice energy) is correct because lattice energy is the energy required to break apart an ionic lattice, opposing the dissolution process.
263 Medium
263 of 560
Remembering +4 | -1
The heat of combustion of a substance is :
View answer and explanation
Correct option: B
Always negative
Option B (Always negative) is correct because combustion reactions release heat, making them exothermic with a negative enthalpy change.
264 Hard
264 of 560
Analyzing +4 | -1
The enthalpy of atomization of PH3(g)\mathrm{PH}_3(\mathrm{g}) is +954 kJ/mol+954 \mathrm{~kJ} /\mathrm{mol} and that of P2H4\mathrm{P}_2 \mathrm{H}_4 is +1.488 MJ/mol.+1.488 \mathrm{~MJ} / \mathrm{mol}. The bond energy of the P−P\mathrm{P}-\mathrm{P} bond is
View answer and explanation
Correct option: C
216 kJ/mol216 \mathrm{~kJ} / \mathrm{mol}
Option C (216 kJ/mol) is correct because the bond energy of the P-P bond is calculated by subtracting the energy required to break the P-H bonds in P2H4 from the total enthalpy of atomization and then accounting for the P-H bonds in PH3.
265 Medium
265 of 560
Application +4 | -1
A person requires 2870 kcal of energy to lead normal daily life. If heat of combustion of cane sugar is -1349 kcal, then his daily consumption of sugar is:
View answer and explanation
Correct option: A
728 g
Option A (728 g) is correct because the daily sugar consumption is calculated by dividing the required energy by the energy per gram of sugar.
266 Easy
266 of 560
Remembering +4 | -1
The work done by a system in an expansion against a constant external pressure is:
View answer and explanation
Correct option: B
−P.ΔV- P.\Delta V
Option B is correct because the work done by a system during expansion against a constant external pressure is given by -PΔV.
267 Easy
267 of 560
Remembering +4 | -1
A thermodynamic state function is:
View answer and explanation
Correct option: C
A quantity whose value is independent of the path
Option C is correct because a thermodynamic state function's value depends only on the initial and final states, not the path taken.
268 Medium
268 of 560
Application +4 | -1
Heat of formation of H2O{H_2}O is -188 kJ /mol and H2O2{H_2}{O_2} is -286 kJ/mol.The enthalpy change for the reaction 2H2O2→2H2O+O22{H_2}{O_2} \to 2{H_2}O + {O_2} is
View answer and explanation
Correct option: A
196 kJ
Option A (196 kJ) is correct because it is calculated using Hess's Law, considering the enthalpy of formation of reactants and products.
269 Medium
269 of 560
Application +4 | -1
The work done during the expansion of a gas from a volume of 4 dm34\,d{m^3} to 6 dm36\,d{m^3} against a constant external pressure of 3 atm is
View answer and explanation
Correct option: B
− 608J- \,608J
Option D (− 608J- \,608J) is correct because it correctly calculates the work done against a constant external pressure using the formula W=−PextΔVW = -P_{ext} \Delta V.
270 Medium
270 of 560
Application +4 | -1
1 mole of an ideal gas at 25℃ is subjected to expand reversibly ten times of its initial volume. The change in entropy of expansion is:
View answer and explanation
Correct option: A
19.15JK−1mol−119.15\mathrm{J}{\mathrm{K}^{ - 1}}\mathrm{mo}{\mathrm{l}^{ - 1}}
Option A (19.15JK−1mol−119.15\mathrm{J}{\mathrm{K}^{ - 1}}\mathrm{mo}{\mathrm{l}^{ - 1}}) is correct because the change in entropy for reversible expansion is calculated using ΔS=nRln⁡(V2V1)\Delta S = nR \ln(\frac{V_2}{V_1}).
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