Test: Single Correct MCQs: Thermodynamics | JEE Advanced
16 Questions MCQ Test Chemistry for JEE | Test: Single Correct MCQs: Thermodynamics | JEE Advanced
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The difference between heats of reaction at constant pressure and constant volume for the reaction : 2 C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l) at 25ºC in kJ is (1991)
TIPS/Formulae : Heat capacity at constant volume (qv) = ΔE Heat capacity of constant pressure (qp) = ΔH
ΔH = ΔE +ΔnRT or ΔH – ΔE = ΔnRT
Δn = no. of moles of gaseous products – no. of moles of gaseous reactants
= 12 – 15 = –3
ΔH – ΔE = – 3 × 8.314 × 298 J = – 7.43 kJ.
For which change ΔH ≠ ΔE : (1995S)
TIPS/Formulae : ΔH = ΔE + ΔnRT For ΔH ≠ ΔE, Δn ≠ 0
Where Δn = no. of moles of gaseous products – no. of moles of gaseous reactants
(a) Δn = 2 – 2 = 0
(b) Δn = 0 (∵ they are either in solid or liquid state)
(c) Δn = 1– 1= 0 (∵ C is in solid state)
(d) Δn = 2 – 4 = – 2
∴ (d) is correct answer
The ΔH0f for CO2(g), CO(g) and H2O(g) are –393.5, –110.5 and –241.8 kJ mol–1 respectively. The standard enthalpy change (in kJ) for the reaction CO2(g) + H2(g) → CO(g) + H2O(g) is (2000S)
CO2(g) + H 2(g) ¾¾® CO(g)+ H2O(g) , ΔH = ?
Given, ΔHf CO2(g) = –393.5 kJ/mol
ΔHf CO(g) = – 110.5 kJ/mol
ΔHf H2O2(g) = – 241.8 kJ/mol
∴ ΔH = [ΔHf CO( g) +ΔHf H2O(g)]
– [ΔHf CO2(g) +ΔHf H2(g)]
= [– 110.5 + (– 241.8)] – (– 393.5 + 0)
= 41.2 kJ mol–1
In thermodynamics, a process is called reversible when (2001S)
In a reversible process, the driving and the opposite forces are nearly equal, hence the system and the surroundings always remain in equilibrium with each other.
Which one of the following statements is false? (2001S)
Work is not a state function because it depends upon the path followed.
One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0L, 95(K) → (4.0 atm, 5.0 L, 245K) with a change in internal energy, ΔU= 30.0L atm . The change in enthalpy (ΔH) of the process in L atm is (2002S)
TIPS/Formulae : ΔH = ΔU +P2V2 – P1V1 Given, ΔU = 30.0 L atm
P1 = 2.0 atm, V1 = 3.0 L, T1 = 95 K P2 = 4.0 atm,
V2 = 5.0 L, T2 = 245 K ΔH = ΔU + P2V2 – P1V1
= 30 + (4 × 5) – (2 × 3) = 30 + 20 – 6 = 44 L atm.
Which of the reaction defines ΔH°f ? (2003S)
TIPS/Formulae : ΔH°f is the enthalpy change when 1 mole of the substance is formed from its elements in their standard states.
In (a) carbon is present in diamond however standard state of carbon is graphite. Again, in (d) CO (g) is involved so it can’t be the right option. Further in (c) 2 moles of NH3 are generated. Hence the correct option is (b).
Two moles of an ideal gas is expanded isothermally and reversibly from 1 litre to 10 litre at 300 K. The enthalpy change (in kJ) for the process is (2004S)
ΔH = nCp ΔT solution; since ΔT = 0 so, ΔH = 0
The enthalpy of vapourization of liquid is 30 kJ mol–1 and entropy of vapourization is 75 J mol–1 K. The boiling point of the liquid at 1 atm is (2004S)
∴ T = 400 K
The direct conversion of A to B is difficult, hence it is carried out by the following shown path :
Given 
where e.u. is the entropy unit, then ΔS(A→B) is (2006 - 3M, –1)
The value of log10 K for a reaction A 
B is (Given : 1 ΔrH°298K =-54.07 kJ mol-1, ΔrS°298K = 10 JK–1 mol–1 and R = 8.314 JK–1 mol–1; 2.303 × 8.314 × 298 = 5705) (2007)
For the process H2O(l) (1 bar, 373 K) → H2O(g) (1 bar, 373 K), the correct set of thermodynamic parameters is (2007)
Since, liquid is passing in to gaseous phase so entropy will increase and at 373 K during the phase transformation it remains at equilibrium. So, ΔG = 0.
The species which by definition has ZERO standard molar enthalpy of formation at 298 K is (2010)
The species in its elemental form has zero standard molar enthalpy of formation at 298 K. At 298K, Cl2 is gas while Br2 is liquid.
The standard enthalpies of formation of CO2(g), H2O(l) and glucose(s) at 25°C are –400 kJ/mol, –300 kJ/mol and –1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25°C is (JEE Advanced 2013-I)
The standard enthalpy of the combustion of glucose can be calculated by the eqn.
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
ΔHC = 6 × ΔHf(CO2) + 6 × ΔHf (H2O) – ΔHf [C6H12O6]
ΔH° = 6 (–400) + 6(–300) – (–1300) = –2900 kJ/mol
For one gram of glucose, enthalpy of combustion
- 16.11 kJ / gm
For the process (JEE Adv. 2014) H2O(l) → H2O(g) at T = 100°C and 1 atmosphere pressure, the correct choice is
Given conditions are boiling conditions for water due to which
ΔStotal = 0
ΔSsystem + ΔSsurroundings = 0
ΔSsystem = – ΔSsurroundings
For process, ΔSsystem > 0 ΔSsurroundings < 0
One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings (ΔSsurr) in JK–1 is (1 L atm = 101.3 J) (JEE Adv. 2016)
From 1st law of thermodynamics
qsys = ΔU – w = 0 – [–Pext.ΔV]
= 3.0 atm × (2.0 L – 1.0 L) = 3.0 L-atm
∴ 
= –1 .01 3 J/K
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