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NEET • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Mixed difficulty

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361 Easy
361 of 560
Application +4 | -1
The enthalpy of vaporisation of a substance is 840 J/mol and its boiling point is -173℃. Its entropy of vaporisation is
View answer and explanation
Correct option: A
4.8 J/mol/K
Option A (4.8 J/mol/K) is correct because the entropy of vaporization is calculated by dividing the enthalpy of vaporization by the boiling point in Kelvin.
362 Easy
362 of 560
Remembering +4 | -1
What are the units of entropy?
View answer and explanation
Correct option: B
cal. K−1\mathrm{K}^{-1}
Option B (cal. K−1\mathrm{K}^{-1}) is correct because entropy is defined as the heat absorbed divided by the temperature, hence the units.
363 Medium
363 of 560
Understanding +4 | -1
Which of the following statements is incorrect?
View answer and explanation
Correct option: D
Both (1) and (2)
Option C (Both H{H} and q{q} are state functions) is correct because heat (q) is a path function, not a state function, and therefore the statement is incorrect. H (enthalpy) is a state function.
364 Easy
364 of 560
Understanding +4 | -1
The temperature of the system increases during an:
View answer and explanation
Correct option: B
Adiabatic compression
Option B is correct because during adiabatic compression, work is done on the system, increasing its internal energy and thus its temperature.
365 Hard
365 of 560
Analyzing +4 | -1
When 1 mole of Na(s)\mathrm{Na}(\mathrm{s}) is dissolved in large volume of water at 298 K298 \mathrm{~K} and 1 bar,184 kJ/mol1 \mathrm{~bar}, 184 \mathrm{~kJ} / \mathrm{mol} heat is released. When 1 mole of Na2O(s)\mathrm{Na}_2 \mathrm{O}(\mathrm{s}) is dissolved in large volume of water at 298 K298 \mathrm{~K} and 1 bar,238 kJ/mol1 \mathrm{~bar}, 238 \mathrm{~kJ} / \mathrm{mol} is released. If the enthalpy of formation of water is −286 kJ/mol,-286 \mathrm{~kJ} / \mathrm{mol}, then the enthalpy of formation of sodium oxide is
View answer and explanation
Correct option: C
−416 kJ/mol-416 \mathrm{~kJ} / \mathrm{mol}
Option C (-416 kJ/mol) is correct because the enthalpy of formation of sodium oxide can be calculated using Hess's Law and the given enthalpies of dissolution and formation of water.
366 Hard
366 of 560
Application +4 | -1
One mole of a perfect gas expands isothermally and reversibly from 10 dm310 \mathrm{~dm}^3 to 20 dm320 \mathrm{~dm}^3 at 300 K300 \mathrm{~K}. Δu,q\Delta u, q and work done in the process are [R=8.3×10−3   kJK−1 mol−1]\left[\mathrm{R}={8 . 3} \times {1 0}^{-{3}}\,\mathrm\,\,{kJ} \mathrm{K}^{-1} \mathrm{~mol}^{-{1}}\right]
View answer and explanation
Correct option: A
0,1.726 kJ,−1.726 kJ0,1.726 \mathrm{~kJ},-1.726 \mathrm{~kJ}
Option A (0, 1.726 kJ, -1.726 kJ) is correct because in an isothermal process, the change in internal energy is zero, and the heat absorbed is equal to the negative of the work done.
367 Medium
367 of 560
Understanding +4 | -1
The heat of formation (∆Hf)ofH2O(l)({∆H_f})of{H_2}O(l)is equal to:
View answer and explanation
Correct option: C
Molar heat of combustion of H2(g){\mathrm{H}_2}\left( \mathrm{g} \right)
Option C is correct because the heat of formation of H2O(l) is defined as the molar heat of combustion of H2(g).
368 Easy
368 of 560
Remembering +4 | -1
Second law of thermodynamics defines
View answer and explanation
Correct option: A
Entropy
Option A is correct because the second law of thermodynamics introduces the concept of entropy as a measure of disorder.
369 Easy
369 of 560
Analyzing +4 | -1
Enthalpy of neutralization of the reaction between CH3COOH(aq)\mathrm{CH}_3 \mathrm{COOH} \text{(aq)} and NaOH(aq)\mathrm{NaOH} \text{(aq)} is −13.2 kcal/eq-13.2 \mathrm{~kcal/eq} and that of the reaction between H2SO4(aq)\mathrm{H}_2 \mathrm{SO}_4(\mathrm{aq}) and KOH(aq)\mathrm{KOH}(\mathrm{aq}) is −13.7 kcal/eq.-13.7 \mathrm{~kcal} / \mathrm{eq}. The enthalpy of dissociation of CH3COOH(aq)\mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq}) is
View answer and explanation
Correct option: B
+0.5 kcal eq−1+0.5 \mathrm{~kcal} \mathrm{~eq}^{-1}
Option B (+0.5 kcal/eq-1) is correct because it is the difference between the standard enthalpy of neutralization and the experimental enthalpy of neutralization.
370 Hard
370 of 560
Application +4 | -1
The H─H bond energy is 430 kJ mol−1mo{l^{ - 1}} and Cl─Cl bond energy is 240 kJ mol−1mo{l^{ - 1}} . ∆H for HCl is -90 kJ. The H─Cl bond energy is about:
View answer and explanation
Correct option: A
425kJmol−1425kJmo{l^{ - 1}}
Option A (425kJmol−1425kJmo{l^{ - 1}}) is correct because it is calculated using the formula: ∆H = Σ(Bond energies of reactants) - Σ(Bond energies of products).
371 Medium
371 of 560
Remembering +4 | -1
The term −TΔStotal-T \Delta S_{\text {total}} at constant temperature and pressure is known as
View answer and explanation
Correct option: A
Gibb's function
Option A (Gibb's function) is correct because the term −TΔStotal-T \Delta S_{\text {total}} at constant temperature and pressure is defined as the Gibbs free energy function.
372 Medium
372 of 560
Remembering +4 | -1
For the process to occur under adiabatic conditions, the correct condition is
View answer and explanation
Correct option: C
q=0q=0
Option C (q=0q=0) is correct because an adiabatic process is defined as a process in which no heat is exchanged between the system and its surroundings.
373 Medium
373 of 560
Analyzing +4 | -1
Four monobasic acids A,B,C and D have their respective ΔneutH0{\Delta _{neut}}{H^0} values are -11.5 , -7.5, -12.4 and -8.9 kcal/mol. Which of the following acids has the highest pKa value ?
View answer and explanation
Correct option: B
B
Option B (B) is correct because a higher pKa value indicates a weaker acid, and a weaker acid has a less negative enthalpy of neutralization.
374 Medium
374 of 560
Application +4 | -1
16 g oxygen gas expands at STP to occupy double of its original volume. The work done during the process is:
View answer and explanation
Correct option: D
272.8 kcal
Option D (272.8 kcal) is correct because it represents the work done during the isothermal expansion of oxygen gas at STP to double its volume, calculated using the formula W = -nRTln(V2/V1).
375 Easy
375 of 560
Application +4 | -1
For strong acid strong base neutralisation energy for 1 mole H2O{H}_2 {O} formation is −57.1 kJ-57.1 \mathrm{~kJ}. If 0.250.25 mole of strong monoprotic acid is reacted with 0.50.5 mole of strong base then enthalpy of neutralisation is
View answer and explanation
Correct option: A
−(0.25×57.1)-(0.25 \times 57.1)
Option A (-(0.25 x 57.1)) is correct because the enthalpy of neutralization is proportional to the moles of water formed, which is limited by the limiting reactant (acid).
376 Medium
376 of 560
Understanding +4 | -1
Which is not characteristic of thermochemical equation?
View answer and explanation
Correct option: D
It indicates whether reaction would occur or not
Option D (It indicates whether reaction would occur or not) is correct because a thermochemical equation does not predict the spontaneity or feasibility of a reaction, only the enthalpy change.
377 Hard
377 of 560
Application +4 | -1
A cylinder of gas is assumed to contain 11.2 kg of butane (C4H10)\mathrm{(}{\mathrm{C}_4}{\mathrm{H}_{10}}). If a normal family needs 20000 kJ of energy per day. The cylinder will last (Given that ∆H for combustion of butane is -2658 kJ)
View answer and explanation
Correct option: C
26 days
Option C (26 days) is correct. It is calculated by determining the total energy available from the butane and dividing it by the daily energy requirement.
378 Medium
378 of 560
Remembering +4 | -1
If a chemical change is brought about by one or more methods in one or more steps, then the amount of heat absorbed or evolved during the complete course of reaction is same, whichever method was followed. This law is known as :
View answer and explanation
Correct option: B
Hess’s law
Option B (Hess’s law) is correct because Hess's law states that the enthalpy change of a reaction is independent of the pathway.
379 Medium
379 of 560
Analyzing +4 | -1
Assertion (A): The dissolution of NH4Cl\mathrm{NH}_4 \mathrm{Cl} in water is endothermic but it dissolves in water.
Reason (R): On dissolution of NH4Cl\mathrm{NH}_4 \mathrm{Cl}, entropy increases.
View answer and explanation
Correct option: B
Both Assertion and Reason are correct; Reason is not the correct explanation of Assertion
Option B (Both Assertion and Reason are correct; Reason is not the correct explanation of Assertion) is correct. NH4Cl dissolution is endothermic and spontaneous due to increased entropy, but the entropy increase isn't the *only* reason for dissolution.
380 Medium
380 of 560
Analyzing +4 | -1
The standard reaction Gibbs energy for a chemical reaction at an absolute temperature TT is given by, ΔrG∘=A−BT{\Delta _r}{G^ \circ } = A - BT Where AA and BB are non-zero constants. Which of the following is true about this reaction?
View answer and explanation
Correct option: D
Endothermic if, A>0A > 0
Option D (Endothermic if, A>0A > 0) is correct because if A > 0, then at T=0, ΔrG∘=A>0\Delta_r G^\circ = A > 0. As temperature increases, the reaction becomes more spontaneous if B is positive, but the initial value is endothermic.
381 Medium
381 of 560
Remembering +4 | -1
ΔG\Delta G for a spontaneous reaction is
View answer and explanation
Correct option: B
negative
Option B (negative) is correct because a spontaneous reaction is characterized by a decrease in Gibbs free energy, resulting in a negative Δ G.
382 Hard
382 of 560
Analyzing +4 | -1
One mole of an ideal gas undergoes a reversible process in which the entropy of the gas changes with absolute temperature TT as S=aT+Cv,mS=a T+C_{\mathrm{v}, \mathrm{m}} ln⁡T,\ln T, where aa is a positive constant. If T=T0T=T_0 at V=V0,V=V_0, then the volume dependence of the gas on temperature in this process is
View answer and explanation
Correct option: C
T=T0+Ra⋅ln⁡VV0T=T_{0}+\frac{R}{a} \cdot \ln \frac{V}{V_{0}}
Option C is correct because it represents the derived relationship between temperature and volume based on the given entropy equation and thermodynamic principles.
383 Medium
383 of 560
Analyzing +4 | -1
Under the same conditions how many mL of 1M KOH and 0.5 M H2SO4{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} solutions, respectively when mixed for a total volume of 100 mL produce the highest rise in temperature?
View answer and explanation
Correct option: D
    50:50\;\;50\mathrm{ }:\mathrm{ }50
Option D (50 : 50) is correct because the highest rise in temperature occurs when the acid and base completely neutralize each other, which happens at the stoichiometric ratio.
384 Medium
384 of 560
Understanding +4 | -1
The enthalpy change is negative for :
View answer and explanation
Correct option: A
Cl−(g)+aq→Cl−(aq)\mathrm{C}{\mathrm{l}^ - }\left( \mathrm{g} \right) + \mathrm{aq} \to \mathrm{C}{\mathrm{l}^ - }\left( {aq} \right)
Option A (Cl−(g)+aq→Cl−(aq)\mathrm{C}{\mathrm{l}^ - }\left( \mathrm{g} \right) + \mathrm{aq} \to \mathrm{C}{\mathrm{l}^ - }\left( {aq} \right)) is correct because the hydration of ions is an exothermic process, releasing energy and resulting in a negative enthalpy change.
385 Hard
385 of 560
Analyzing +4 | -1
An ideal gas with the adiabatic exponent γ\gamma goes through a process P=P0−αV,P=P_{0}-\alpha V, where P0P_{0} and α\alpha are positive constants and VV is the volume. At what volume will the gas entropy have the maximum value?
View answer and explanation
Correct option: B
γ⋅P0α(γ+1)\frac{\gamma \cdot P_0}{\alpha(\gamma+1)}
Option B (γ⋅P0α(γ+1)\frac{\gamma \cdot P_0}{\alpha(\gamma+1)}) is correct because by using the given equation and thermodynamic relations, the volume at which entropy is maximum can be derived.
386 Easy
386 of 560
Remembering +4 | -1
Unit of entropy is
View answer and explanation
Correct option: A
JK−1mol−1J{K^{ - 1}}mo{l^{ - 1}}
Option A is correct. Entropy is measured in Joules per Kelvin per mole (J/K·mol), representing the change in energy with respect to temperature and the amount of substance.
387 Medium
387 of 560
Application +4 | -1
Under isothermal condition for one mole of ideal gas what is the ratio of work done under reversible to irreversible process, initially held at 20 atm20 {~atm} undergoes expansion from 1 L1 {~L} to 2 L2 {~L}, at 298 K298 {~K}, under external pressure of 10 atm10 {~atm} ?
View answer and explanation
Correct option: A
1.7
Option A (1.7) is correct because the ratio of work done in a reversible isothermal process to an irreversible process is calculated using the respective work equations.
388 Medium
388 of 560
Understanding +4 | -1
The first law of thermodynamic is expressed as
View answer and explanation
Correct option: C
q=ΔE−Wq = \Delta E - W
The correct answer is B (ΔE=q−W\Delta E = q - W) because it accurately represents the first law of thermodynamics, stating that the change in internal energy equals heat added to the system minus work done *by* the system.
389 Medium
389 of 560
Application +4 | -1
A system absorbs 600 J of heat and work equivalent to 300 J on its surroundings. The change in internal energy is
View answer and explanation
Correct option: A
300 J
Option A (300 J) is correct because the change in internal energy is the heat absorbed minus the work done by the system (ΔU = Q - W).
390 Medium
390 of 560
Remembering +4 | -1
ΔG\Delta \mathrm{G} is equal to
View answer and explanation
Correct option: B
ΔG∘+RTln⁡Q\Delta G^{\circ}+R T \ln Q
Option B (ΔG∘+RTln⁡Q\Delta G^{\circ}+R T \ln Q) is correct because it represents the relationship between Gibbs free energy change, standard Gibbs free energy change, and the reaction quotient.
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