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NEET • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Mixed difficulty

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271 Hard
271 of 560
Application +4 | -1
2 mole of an ideal gas undergo isothermal and reversible expansion from 2 litre to 20 liltre at 127∘C127^{\circ} \mathrm{C} the work done by the gas is
View answer and explanation
Correct option: B
−15.3 kJ-15.3 \mathrm{~kJ}
Option B (−15.3 kJ-15.3 \mathrm{~kJ}) is correct because it correctly applies the formula for isothermal and reversible expansion, W=−nRTln⁡(V2/V1)W = -nRT \ln(V_2/V_1).
272 Medium
272 of 560
Remembering +4 | -1
In thermodynamics, a quantity whose value
simply depends upon the initial and final states
of the system is called
View answer and explanation
Correct option: B
a state function
Option B (a state function) is correct because a state function's value depends only on the initial and final states, not the path taken.
273 Medium
273 of 560
Understanding +4 | -1
Heat energy change during the chemical reaction, CO+12O2→CO2\mathrm{CO} + \frac{1}{2}{\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2}is known as :
View answer and explanation
Correct option: A
Heat of combustion of CO
Option A (Heat of combustion of CO) is correct because the reaction describes the complete burning of CO in oxygen to form CO2, which is the definition of heat of combustion.
274 Medium
274 of 560
Understanding +4 | -1
A thermally isolated gaseous system can exchange energy with the surroundings. The mode of transference of energy can be:
View answer and explanation
Correct option: B
Work
Option B (Work) is correct because a thermally isolated system cannot exchange heat with the surroundings, but it can still exchange energy through work.
275 Easy
275 of 560
Understanding +4 | -1
For the reaction, N2(g)+3H2(g)→2NH3(g),{\mathrm{N}_2}\left( \mathrm{g} \right) + 3{\mathrm{H}_2}\left( \mathrm{g} \right) \to 2\mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right), which is true?
View answer and explanation
Correct option: B
∆H< ∆U
Option B, ∆H< ∆U, is correct because for the given reaction, the number of moles of gas decreases, leading to a negative value for Δng\Delta n_g, and thus ∆H< ∆U.
276 Medium
276 of 560
Remembering +4 | -1
Work done in reversible adiabatic process is given by:
View answer and explanation
Correct option: B
nR(γ−1)(T2−T1)\frac{{nR}}{{\left( {\gamma - 1} \right)}}\left( {{T_2} - {T_1}} \right)
Option B (nR(γ−1)(T2−T1)\frac{{nR}}{{\left( {\gamma - 1} \right)}}\left( {{T_2} - {T_1}} \right)) is correct because this formula represents the work done in a reversible adiabatic process, relating it to the number of moles, gas constant, adiabatic index, and temperature change.
277 Medium
277 of 560
Remembering +4 | -1
The van&rsquo;t Hoff reaction isotherm is:
View answer and explanation
Correct option: B
−∆G=RT log⁡eKp-∆ G = RT\,{\log _e}{K_p}
Option B (-∆G = RT logₑ Kₚ) is correct because it represents the relationship between Gibbs free energy change and the equilibrium constant.
278 Medium
278 of 560
Analyzing +4 | -1
Given are two statements one is labelled as Assertion A and other is labelled as Reason R.
Assertion A: Magnesium can reduce Al2O3A{l_2}{O_3} at a temperature below 1350∘ C1350^\circ \,C, while above 1350∘ C1350^\circ \,C aluminium can reduce MgOMgO.
Reason R: The melting and boiling points of magnesium are lower than those of aluminium.
In light of the above statements, choose most appropriate answer from the options given below:
View answer and explanation
Correct option: B
Both A and R are correct, but R is NOT the correct explanation of A.
Option B is correct because both the assertion and the reason are true, but the reason does not explain the assertion. The Ellingham diagram explains the assertion.
279 Medium
279 of 560
Application +4 | -1
Given, C+O2→CO2+94.2kcal….(i)\mathrm{C} + {\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2} + 94.2\mathrm{kcal} \ldots \mathrm{.}\left( \mathrm{i} \right)H2+12O2→H2O+68.3kcal….(ii){\mathrm{H}_2} + \frac{1}{2}{\mathrm{O}_2} \to {\mathrm{H}_2}\mathrm{O} + 68.3\mathrm{kcal} \ldots .\left( {\mathrm{ii}} \right)CH4+2O2→CO2+2H2O+210.8kcal….(iii)\mathrm{C}{\mathrm{H}_4} + 2{\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O} + 210.8\mathrm{kcal} \ldots \mathrm{.}\left( {\mathrm{iii}} \right)The heat of formation in kcal will be :
View answer and explanation
Correct option: C
-20
Option C (-20) is correct because it correctly calculates the heat of formation of methane by using Hess's Law and the given combustion reactions.
280 Hard
280 of 560
Understanding +4 | -1
The heat change for a reaction CO(g)+12O2(g) →  CO2(g)CO\left( g \right) + \frac{1}{2}{O_2}\left( g \right)\,\xrightarrow{\,}\,C{O_2}\left( g \right) refers to
View answer and explanation
Correct option: D
Enthalpy of combustion of COCO
The heat change for the reaction represents the enthalpy of combustion of CO because CO is reacting with oxygen to form CO2.
281 Hard
281 of 560
Application +4 | -1
The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0℃ is:
View answer and explanation
Correct option: A
5.260 cal/(mol K)
Option A (5.260 cal/(mol K)) is correct because it is calculated by dividing the enthalpy of fusion by the temperature at which melting occurs.
282 Easy
282 of 560
Application +4 | -1
For the melting of NaClNaCl heat required is 7.26 K cal mol−17.26\,K\,cal\,mo{l^{ - 1}} and ΔS\Delta S increases by 6.73 cal mol−1.K−1.6.73\,cal\,mo{l^{ - 1}}.{K^{ - 1}}. The melting point of the salt is
View answer and explanation
Correct option: A
805.75∘C805.75^\circ C
Option A (805.75∘C805.75^\circ C) is correct because the melting point is calculated using the formula T=ΔHΔST = \frac{{\Delta H}}{{\Delta S}} and then converted from Kelvin to Celsius.
283 Hard
283 of 560
Understanding +4 | -1
If we take water at temperature TAT_A in a container having thermally conducting walls and enclosed in a huge heat reservoir at temperature TBT_B. What is the correct expression of change in internal energy if no work is done at constant volume?
View answer and explanation
Correct option: B
ΔU=q\Delta U=q
Option B (ΔU=q\Delta U=q) is correct because, at constant volume, the change in internal energy is equal to the heat absorbed or released by the system when no work is done.
284 Medium
284 of 560
Analyzing +4 | -1
Statement - I: In an adiabatic expansion of an ideal gas, the entropy of the surrounding
remains unchanged independent whether the expansion is reversible or irreversible.

Statement - II: In an adiabatic process, the surrounding does not exchange heat with the
system and the process may be reversible or irreversible for system but it always remain
reversible for the sorrounding.
View answer and explanation
Correct option: A
If both Statement-I and Statement-II are true, and Statement-II is the correct
explanation of Statement-I
Option A is correct because both statements are true, and Statement-II correctly explains why the entropy of the surrounding remains unchanged in an adiabatic expansion.
285 Easy
285 of 560
Remembering +4 | -1
Which of the following is an example of extensive property?
View answer and explanation
Correct option: C
Mass
Option C (Mass) is correct because mass is an extensive property, meaning it depends on the amount of substance, while temperature, density, and pressure are intensive properties.
286 Medium
286 of 560
Application +4 | -1
The heat change taking place during the reaction   H2O(l)\;{\mathrm{H}_2}\mathrm{O}\left( l \right)→H2O(g)\to {\mathrm{H}_2}\mathrm{O(g)} is: [Given,ΔHf\Delta {H_f} of H2O(g){\mathrm{H}_2}\mathrm{O}(g) = -57 kcal, ΔHf=H2O(l)\Delta {H_f} = {\mathrm{H}_2}\mathrm{O}\left( l \right) = -68.3 kcal]
View answer and explanation
Correct option: A
+11.3kcal+ 11.3\mathrm{kcal}
Option A (+11.3 kcal) is correct. The heat change is the enthalpy of vaporization, calculated as the difference between the enthalpy of formation of gaseous water and liquid water.
287 Easy
287 of 560
Application +4 | -1
For the reaction;   C2H5OH(l)+3O2(g)→  2CO2(g)+3H2O(l)\;{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}\left( l \right) + 3{\mathrm{O}_2}\left( \mathrm{g} \right) \to \;2C{O_2}(g) + 3{H_2}O(l) which one is true?
View answer and explanation
Correct option: A
∆H= ∆U-RT
Option D (∆H= ∆U-RT) is correct because ∆H = ∆U + ∆n_g RT, and in this reaction, ∆n_g = (2) - (3) = -1.
288 Easy
288 of 560
Understanding +4 | -1
A gas is allowed to expand under reversible adiabatic conditions then :-
View answer and explanation
Correct option: C
ΔS=0\Delta {S}=0
Option C (ΔS=0\Delta {S}=0) is correct because in a reversible adiabatic process, the change in entropy is zero.
289 Easy
289 of 560
Application +4 | -1
What is the entropy change in JK−1\mathrm{JK}^{-1} during the melting of 27.3 g27.3 \mathrm{~g} of ice at 0∘C0^{\circ} \mathrm{C} ? (Latent heat of fusion of ice =330Jg−1=330 \mathrm{Jg}^{-1} )
View answer and explanation
Correct option: D
33
Option D (33) is correct because the entropy change is calculated by dividing the heat absorbed during melting by the temperature in Kelvin.
290 Medium
290 of 560
Application +4 | -1
How much heat (in kJ) is evolved when 2.7 g of aluminium reacts with a stoichiometric amount of Fe2O3\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}?2Al2+Fe2O3→2Fe+Al2O3;ΔHo=−852 kJ2\mathrm{A}{\mathrm{l}_2} + \mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3} \to 2\mathrm{Fe} + \mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3};\Delta {H^o} = - 852\,\mathrm{kJ}
View answer and explanation
Correct option: C
−42.6 kJ- 42.6\,\mathrm{kJ}
Option C (-42.6 kJ) is correct because it is calculated based on the stoichiometry of the reaction and the given enthalpy change.
291 Medium
291 of 560
Application +4 | -1
A solution of 500 mL of 2M KOH is added to 500 mL of 2M HCl and the mixture is well shaken. The rise in temperature   T1\;{T_1} is noted. The experiment is then repeated using 250 mL of each solution and rise in temperature   T2\;{T_2} is again noted. Assume all heat is taken up by the solution:
View answer and explanation
Correct option: A
  T1\;{T_1}=  T2\;{T_2}
Option A (T1=T2) is correct because the heat released per mole of reaction is the same in both cases, and the total volume of solution is also the same, leading to the same temperature rise.
292 Medium
292 of 560
Application +4 | -1
Heat of combustion ∆H for C(s), H2{\mathrm{H}_2}(g) and CH4\mathrm{C}{\mathrm{H}_4} (g) are -94,-68 and -213 kcal/mole then ∆H for C(s)+2H2(g)→CH4(g)is:\mathrm{C}\left( s \right) + 2{\mathrm{H}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right)\mathrm{i}s:
View answer and explanation
Correct option: A
-17 kcal
Option A (-17 kcal) is correct because it is calculated using Hess's law by manipulating the given heats of combustion.
293 Medium
293 of 560
Understanding +4 | -1
Molar heat capacity is given by :
View answer and explanation
Correct option: D
None of these
The correct answer is None of these because molar heat capacity is defined as the amount of heat required to raise the temperature of one mole of a substance by one degree Celsius, which is mathematically represented as C=dQndTC = \frac{dQ}{n dT}, where n is the number of moles.
294 Hard
294 of 560
Understanding +4 | -1
Consider the following equation, q= coeff. ×ΔTq=\text { coeff. } \times \Delta T. The magnitude of the coefficient depends on
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the coefficient in the equation depends on the size, composition, and nature of the system.
295 Medium
295 of 560
Understanding +4 | -1
The heat of formation is the change in enthalpy accompanying the formation of a substance from its elements at 298 K and 1 atm pressure. Since, the enthalpies of elements in their most stable state are taken to be zero, the heat of formation of compounds is :
View answer and explanation
Correct option: C
Standard heat enthalpy of that compound
Option C (Standard heat enthalpy of that compound) is correct because the heat of formation is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states.
296 Medium
296 of 560
Understanding +4 | -1
To calculate the amount of work done in joules during reversible isothermal expansion of an ideal gas, the volume must be expressed in
View answer and explanation
Correct option: D
Any of these
Option D (Any of these) is correct because the work done calculation depends on the change in volume, and any consistent unit can be used as long as the gas constant is adjusted accordingly.
297 Easy
297 of 560
Remembering +4 | -1
Hess’s law is related to:
View answer and explanation
Correct option: A
Change in heat during a reaction
Option A is correct because Hess's law states that the change in enthalpy during a reaction depends only on the initial and final states, not the path taken.
298 Medium
298 of 560
Understanding +4 | -1
Identify the correct statement for change of Gibb’s energy for a system (ΔGsystem\Delta {G_{\mathrm{system}}}) at constant temperature and pressure:
View answer and explanation
Correct option: D
If ΔGsystem\Delta {G_{\mathrm{system}}}=0, the system has attained equilibrium
Option D (If ΔGsystem\Delta {G_{\mathrm{system}}}=0, the system has attained equilibrium) is correct because at equilibrium, the change in Gibbs free energy is zero.
299 Hard
299 of 560
Application +4 | -1
56 g56 \mathrm{~g} of iron reacts with HCl\mathrm{HCl} in a closed vessel of fixed volume. Again, same amount of Fe\mathrm{Fe} is made to react in an open beaker at 27∘C27^{\circ} \mathrm{C}. The work done (in cal) in both the cases will be respectively
View answer and explanation
Correct option: D
0,−6000,-600
Option D (0, -600) is correct because in a closed vessel, no work is done due to fixed volume, while in an open beaker, work is done against the constant atmospheric pressure.
300 Hard
300 of 560
Analyzing +4 | -1
Heat of neutralization of a strong acid HAH A and a weaker acid HBH B with KOHK O H are -13.7 and −12.7Kcalmol−1-12.7 \mathrm{K} \mathrm{cal} \mathrm{mol}^{-1}. When 1 mole of KOH\mathrm{KOH} is added to a mixture containing 1 mole each of HAH A and HBH B, the heat changes was −13.5K-13.5 \mathrm{K} cal. In what ratio is the base distributed between HAH A and HBH B.
View answer and explanation
Correct option: C
4:14: 1
Option C (4:1) is correct because it represents the ratio in which KOH is distributed between the strong acid HA and the weak acid HB, calculated using the given heats of neutralization.
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