JEE Advanced (Single Correct MCQs): Thermodynamics - JEE MCQ

TIPS/Formulae : Heat capacity at constant volume (q_v) = ΔE Heat capacity of constant pressure (q_p) = 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.

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

CO₂(g) + H ₂(g) ¾¾® CO(g)+ H₂O(g) , ΔH = ?

Given, ΔHf CO₂(g) = –393.5 kJ/mol
ΔH_f CO(g) = – 110.5 kJ/mol
ΔH_f H₂O₂(g) = – 241.8 kJ/mol

∴ ΔH = [ΔH_f CO( g) +ΔH_f H₂O(g)]                          
– [ΔH_f CO₂(g) +ΔH_f H₂(g)]
= [– 110.5 + (– 241.8)] – (– 393.5 + 0)

= 41.2 kJ mol^–1

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.

Work is not a state function because it depends upon the path followed.

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.

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).

ΔH = nC_p ΔT solution; since ΔT = 0 so, ΔH = 0

∴ T = 400 K

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 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 enthalpy of the combustion of glucose can be calculated by the eqn.

C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l)
ΔH_C = 6 × ΔH_f(CO₂) + 6 × ΔH_f (H₂O) – ΔH_f [C₆H₁₂O₆]
ΔH° = 6 (–400) + 6(–300) – (–1300) = –2900 kJ/mol

For one gram of glucose, enthalpy of combustion

 - 16.11 kJ / gm

Given conditions are boiling conditions for water due to which
ΔS_total = 0
ΔS_system + ΔS_surroundings = 0
ΔS_system = – ΔS_surroundings
For process, ΔS_system > 0 ΔS_surroundings < 0

From 1st law of thermodynamics
q_sys = ΔU – w = 0 – [–P_ext.ΔV]          
= 3.0 atm × (2.0 L – 1.0 L) = 3.0 L-atm

∴ 

= –1 .01 3 J/K