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NEET • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Mixed difficulty

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211 Hard
211 of 560
Analyzing +4 | -1
Select the correct set of statements considering initial state to be same and final state at same volume in case of comparison.

(I) Magnitude of Work done on the surrounding in case of infinite step expansion is more than single stage expansion for an isothermal process.

(II) Irreversible work is always greater than reversible work for an isothermal process. (with sign)

(III) On an ideal gas in case of single stage expansion and compression, system as well as surrounding are restored back to their original states.

(IV) If gas in thermodynamic equilibrium is taken from state A to state B, by four successive single stage expansions. Then we can plot maximum 4 points on the P−V\mathrm{P}-\mathrm{V} indicator diagram.
View answer and explanation
Correct option: A
I, II only
Option A (I, II only) is correct. Work done is greater in multistep expansion than single-step, and irreversible work is greater than reversible work (with sign).
212 Medium
212 of 560
Application +4 | -1
The values of ∆H and ∆S for the reaction. C(graphite)+CO2(g)→2CO(g){C_{\left( {\mathrm{graphite}} \right)}} + \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right) \to 2\mathrm{CO}\left( \mathrm{g} \right) Are 170 kJ and 170 JK−1\mathrm{J}{\mathrm{K}^{ - 1}} respectively. this reaction will be spontaneous at:
View answer and explanation
Correct option: D
1110 K
Option D (1110 K) is correct because the reaction becomes spontaneous when T > ∆H/∆S, which calculates to be 1000 K.
213 Medium
213 of 560
Remembering +4 | -1
Internal energy of an ideal gas depends on:
View answer and explanation
Correct option: B
Temperature
Option B (Temperature) is correct because the internal energy of an ideal gas depends only on its temperature, as there are no intermolecular forces.
214 Medium
214 of 560
Application +4 | -1
A carnot engine operates between temperature T and 400 K (T>400 K). If efficiency of engine is 25%, the temperature T is:
View answer and explanation
Correct option: C
533.3 K
Option C (533.3 K) is correct because the efficiency of a Carnot engine is given by η=1−TcTh\eta = 1 - \frac{T_c}{T_h}, where TcT_c is the cold reservoir temperature and ThT_h is the hot reservoir temperature. Solving for ThT_h with the given efficiency yields approximately 533.3 K.
215 Hard
215 of 560
Application +4 | -1
5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV=28  JK−1mol−1,{C_V} = 28\mathrm{ }\,\,J{K^{ - 1}}\mathrm{ }mo{l^{ - 1}}, calculate ΔU\Delta U\mathrm{ } and ΔpV\Delta pV for this process, (R=8.0 JK−1mol−1)(R = 8.0\mathrm{ }\,J{K^{ - 1}}\mathrm{ }mo{l^{ - 1}})
View answer and explanation
Correct option: C
ΔU=14 kJ;Δ(pV)=4 kJ\Delta U = 14\mathrm{ }\,kJ;\Delta (pV) = 4\mathrm{ }\,kJ
Option C (ΔU=14 kJ;Δ(pV)=4 kJ\Delta U = 14\mathrm{ }\,kJ;\Delta (pV) = 4\mathrm{ }\,kJ) is correct because it accurately calculates the change in internal energy and the change in pV using the given values.
216 Hard
216 of 560
Application +4 | -1
If a man takes a diet which gives him energy equal to 9500 kJ9500 \mathrm{~kJ} per day and he expands energy all forms to a total of 12000 kJ12000 \mathrm{~kJ} per day, what is the change in internal energy per day? If the energy lost was stored as sucrose (1632 kJ per 100 g100 \mathrm{~g} ), how many days should it take to loss 1 kg1 \mathrm{~kg} ? Ignore the water loss.
View answer and explanation
Correct option: C
6.5 day
Option C (6.5 day) is correct because it is calculated by first finding the daily change in internal energy and then determining how many days it takes to lose 1 kg of sucrose.
217 Easy
217 of 560
Remembering +4 | -1
The amount of heat measured for a reaction in a bomb calorimeter is
View answer and explanation
Correct option: C
∆E
Option C (∆E) is correct because a bomb calorimeter measures heat at constant volume, which corresponds to the change in internal energy (∆E).
218 Easy
218 of 560
Understanding +4 | -1
Entropy decreases during:
View answer and explanation
Correct option: A
Crystallization of sucrose from solution
Option A (Crystallization of sucrose from solution) is correct because crystallization involves the ordering of molecules into a solid lattice, which decreases the disorder and thus the entropy.
219 Easy
219 of 560
Analyzing +4 | -1
A container has hydrogen and oxygen mixture in ratio of 4∶1 by weight, then:
View answer and explanation
Correct option: A
Entropy of these gases increase
Option A (Entropy of these gases increase) is correct because mixing gases generally leads to an increase in entropy due to the increased disorder.
220 Hard
220 of 560
Understanding +4 | -1
STATEMENT-1 : The standard free energy changes of all spontaneously occuring reactions are negative.

STATEMENT-2 : The standard free energies of the elements in their standard states at 1 bar and 298 K are taken as zero.
View answer and explanation
Correct option: B
If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1
Both statements are true, but the second statement does not explain the first. Spontaneous reactions have negative Gibbs free energy changes, and the standard free energies of elements in their standard states are defined as zero.
221 Medium
221 of 560
Remembering +4 | -1
Which one of the following is used to remove most of plutonium from spent nuclear fuel?
View answer and explanation
Correct option: B
O2F2{O_2}{F_2}
Option B (O2F2{O_2}{F_2}) is correct because oxygen difluoride is a strong oxidizing agent used to remove plutonium as PuF6PuF_6.
222 Medium
222 of 560
Understanding +4 | -1
For which process energy will be absorbed?
View answer and explanation
Correct option: D
Separation of an electron from a neutral atom
Option D (Separation of an electron from a neutral atom) is correct because energy is always required to remove an electron from a neutral atom due to the attractive force between the nucleus and the electron.
223 Medium
223 of 560
Understanding +4 | -1
Which is correct about the heat of combustion?
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because heat of combustion is always exothermic, and its value changes with temperature.
224 Medium
224 of 560
Analyzing +4 | -1
Assertion: When ideal gas expand from P1,V1,T1P_1, V_1, T_1 to P2,V2,T2P_2, V_2, T_2 in multiple steps, work done is high where number of steps are high.
Reason: Work is path function.
View answer and explanation
Correct option: B
If both Assertion and Reason are correct, but
Reason is not the correct explanation of
Assertion.
Option B is correct because both the assertion and reason are true, but the reason doesn't fully explain why more steps lead to more work; it's about approaching reversibility.
225 Hard
225 of 560
Application +4 | -1
If the pressure and density of a diatomic gas (γ=7/5)(\gamma=7 / 5) in adiabatic process changes from (P,d)(P, d) to (P′,d′)\left(P^{\prime}, d^{\prime}\right), where p/p′p / p^{\prime} is 32 then d\mathrm{d} '/ d\mathrm{d} is 32x32 x, find xx ?
View answer and explanation
Correct option: A
4
Option A (4) is correct because it correctly calculates the ratio of densities using the adiabatic process equation and the given pressure ratio.
226 Medium
226 of 560
Application +4 | -1
Minimum work is obtained when 1 kg of … gas expanded under 500 kPa to 200 kPa pressure at 0℃.
View answer and explanation
Correct option: A
Chlorine
Option A (Chlorine) is correct because minimum work is obtained when the gas with the highest molar mass expands, as it requires more energy to expand due to stronger intermolecular forces. Chlorine has the highest molar mass among the given options.
227 Medium
227 of 560
Analyzing +4 | -1
The normal boiling point and ΔHvap \Delta {H}_{\text {vap }} of a liquid ' X{X} ' are 400 K400 {~K} and 40 kJ mol−140 {~kJ} {~mol}^{-1} respectively. Assuming ΔHvap \Delta {H}_{\text {vap }} to be constant, which of the following is correct?
(I) ΔSvap >100JK−1 mol−1\Delta {S}_{\text {vap }}>100 {JK}^{-1} {~mol}^{-1} at 400 K400 {~K} and 0.5 atm0.5 {~atm}
(II) ΔSvap <100JK−1 mol−1\Delta {S}_{\text {vap }} < 100 {JK}^{-1} {~mol}^{-1} at 400 K400 {~K} and 1 atm1 {~atm}
(III) ΔSvap <100JK−1 mol−1\Delta {S}_{\text {vap }} < 100 {JK}^{-1} {~mol}^{-1} at 400 K400 {~K} and 2 atm2 {~atm}
(IV) ΔSvap =100JK−1 mol−1\Delta {S}_{\text {vap }}=100 {JK}^{-1} {~mol}^{-1} at 400 K400 {~K} and 1 atm1 {~atm}
View answer and explanation
Correct option: D
I, III and IV
Option D (I, III and IV) is correct. At the normal boiling point and 1 atm, ΔSvap = ΔHvap/T = 100 J/Kmol. Increasing pressure decreases ΔSvap.
228 Hard
228 of 560
Application +4 | -1
The variation of equilibrium constant with temperature is given below:
TemperatureEquilibrium  constantT1=25∘CK1=10T2=100∘CK2=100\begin{array}{cccccccccccccccccccc} {\mathrm{Temperature}}&{\mathrm{Equilibrium\,\, constant}}\\ {{T_1} = {{25}^ \circ }C}&{{K_1} = 10}\\ {{T_2} = {{100}^ \circ }C}&{{K_2} = 100} \end{array}
The values of ΔH∘,ΔG∘    at  T1 and  ΔG∘  at  T2  (in  kJ  mol−1)\Delta \mathrm{H}^\circ ,\Delta \mathrm{G}^\circ \,\,\,\,\mathrm{at }\,\,{\mathrm{T}_\mathrm{1}}\mathrm{ }\,\mathrm{and}\,\,\mathrm{ }\Delta G^\circ \,\,\mathrm{ at}\,\,\mathrm{ }{\mathrm{T}_2}\mathrm{ }\,\,\mathrm{(in}\,\,\mathrm{ kJ}\,\,\mathrm{ mo}{\mathrm{l}^{ - 1}}) respectively, are close to [Use R=8.314JK−1mol−1]R\mathrm{ } = \mathrm{ }8.314\mathrm{ }J{K^{ - 1}}mo{l^{ - 1}}]
View answer and explanation
Correct option: C
28.4, -5.71 and -14.29
Option C (28.4, -5.71 and -14.29) is correct because it accurately calculates ΔH∘\Delta H^\circ using the van't Hoff equation and ΔG∘\Delta G^\circ at both temperatures using ΔG∘=−RTln⁡K\Delta G^\circ = -RT\ln K.
229 Medium
229 of 560
Remembering +4 | -1
Specific heat is defined as:
View answer and explanation
Correct option: A
Heat capacity/g
Option A (Heat capacity/g) is correct because specific heat is defined as the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius.
230 Medium
230 of 560
Application +4 | -1
H2{H_2} (g)+Cl(g)=2HCl(g); ∆H(298 K)=22.06 kcal. For this reaction, ∆U is equal to:
View answer and explanation
Correct option: D
−22.06kcal- 22.06\mathrm{kcal}
Option D (-22.06 kcal) is correct because for the given reaction, the change in the number of moles of gas is zero, therefore ∆U = ∆H.
231 Medium
231 of 560
Analyzing +4 | -1
48 g of C (diamond) on complete combustion evolves 1584 kJ of heat. The standard heat of formation of gaseous carbon is 725 kJ/mol. The energies required for the process
(i)C (graphite) → C(gas)
(ii)C (diamond) → C(gas)are:
View answer and explanation
Correct option: C
725, 723
Option C (725, 723) is correct because it calculates the energy required for sublimation of graphite and diamond using the heat of combustion and heat of formation.
232 Medium
232 of 560
Remembering +4 | -1
The word standard in molar enthalpy change implies
View answer and explanation
Correct option: C
Any temperature and pressure 1 bar
Option C (Any temperature and pressure 1 bar) is correct because the standard condition for molar enthalpy change is defined at 1 bar pressure, while temperature is not standardized.
233 Medium
233 of 560
Application +4 | -1
The heat of combustion of ethanol determined by a bomb calorimeter is - 670.48 kcalmol−1mo{l^{ - 1}} at 250C{25^0}C.What is ∆U at 250C{25^0}C for the following reaction? C2H5OH(l)+72O2(g)→2CO2(g)+3H2O(g){\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}\left( l \right) + \frac{7}{2}{\mathrm{O}_2}\left( \mathrm{g} \right) \to 2\mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right) + 3{\mathrm{H}_2}\mathrm{O}\left( \mathrm{g} \right)
View answer and explanation
Correct option: C
−670.48kcal- 670.48\mathrm{kcal}
Option C (-670.48 kcal) is correct because at constant volume (bomb calorimeter), the heat of combustion is equal to the change in internal energy, ∆U.
234 Easy
234 of 560
Understanding +4 | -1
∆n values in ∆H= ∆U+ ∆nRT may have:
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because ∆n, representing the change in the number of moles of gas, can be an integer, a fraction, or positive/negative depending on the reaction stoichiometry.
235 Medium
235 of 560
Application +4 | -1
For the reaction CCl4(g)+2H2O(g)→CO2(g)+4HCl(g)CC{l_{4\left( g \right)}} + 2{H_2}{O_{\left( g \right)}} \to C{O_{2\left( g \right)}} + 4HC{l_{\left( g \right)}} at constant temperature , ΔH−ΔE\Delta H - \Delta E is
View answer and explanation
Correct option: D
2RT
Option D (2RT) is correct because it correctly calculates the difference between ΔH\Delta H and ΔE\Delta E using the change in the number of moles of gas.
236 Easy
236 of 560
Remembering +4 | -1
Gibbs energy G, enthalpy H and entropy S are related by:
View answer and explanation
Correct option: B
G=H-TS
Option B (G=H-TS) is correct because the Gibbs free energy (G) is defined as G = H - TS, where H is enthalpy, T is temperature, and S is entropy.
237 Medium
237 of 560
Application +4 | -1
When one mole of an ideal gas is compressed to half its initial volume and simultaneously heated to twice in initial temperature, the change in entropy (ΔS)(\Delta S) is
View answer and explanation
Correct option: D
(CV−R)ln⁡2({C_V} - R)\ln 2
Option D (CV−R)ln⁡2({C_V} - R)\ln 2 is correct because the change in entropy is calculated by considering both the volume and temperature changes using the appropriate formula.
238 Easy
238 of 560
Remembering +4 | -1
The total heat content of a system at constant pressure is
View answer and explanation
Correct option: A
Enthalpy
Option A (Enthalpy) is correct because enthalpy is defined as the total heat content of a system at constant pressure.
239 Medium
239 of 560
Understanding +4 | -1
Which of the following is correct for a reaction to be spontaneous?
View answer and explanation
Correct option: B
ΔH=−ve,  ΔS=+ve\Delta H = - ve,\,\,\Delta S = + ve
The correct answer is B (ΔH=−ve, ΔS=+ve\Delta H = - ve,\,\Delta S = + ve) because a reaction is spontaneous when it results in a decrease in enthalpy (negative ∆H) and an increase in entropy (positive ∆S).
240 Medium
240 of 560
Remembering +4 | -1
Entropy change of vaporisation at constant pressure is given by:
View answer and explanation
Correct option: A
ΔS(v)=ΔHvT\Delta {S_{\left( v \right)}} = \frac{{\Delta {H_v}}}{T}
Option A is correct because the entropy change of vaporization at constant pressure is given by ΔS(v) = ΔHv/T.
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