Thermodynamics and Thermochemistry ‌–‌ ‌JEE‌ ‌Advanced‌ ‌Previous‌ ‌Year‌ ‌Questions with Solutions

Thermodynamics and Thermochemistry JEE Advanced Questions with Solutions are available here. We have compiled some of the most important questions on Thermodynamics and Thermochemistry from the past JEE Advanced exams. Therefore, going through the questions and solutions will help JEE aspirants in realizing the question types, patterns and difficulty levels. At the same time, practising solving the questions from this chapter will help them gain better concept clarity as well as they will have enough practice for the exam.

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JEE Advanced Previous Year Questions on Thermodynamics and Thermochemistry

Question 1. 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 (∆Ssurr) in JK^-1 is (1 L atm = 101.3 J)

A. 5.763

B. 1.013

C. -1.013

D. -5.763

Solution: (C)

Since it is an isothermal process, ΔU = 0

dq = −dW = P_ext (V₂ − V₁) = 3L − atm = 3 × 101.3 J

ΔS_surrounding− 3 × 101.3 / 300 J K⁻¹ = −1.013 J K⁻¹

∴ ΔS_surr = −1.013 J K⁻¹

Question 2. Tin is obtained from cassiterite by reduction with coke. Use the data given below to determine the minimum temperature (in K) at which the reduction of cassiterite by coke would take place.

At 298 K: ∆ 0(SnO₂(s)) = −581.0 kJ mol⁻¹, ∆ 0(CO₂(g)) = −394.0 kJ mol⁻¹,

0(SnO₂(s)) = 56.0J K⁻¹mol⁻¹, S^o(Sn(s)) = 52.0 J K⁻¹mol⁻¹,

0(C(s)) = 6.0J K⁻¹mol⁻¹, 0(CO₂(g)) = 210.0 J K⁻¹mol⁻¹.

Assume that the enthalpies and the entropies are temperature independent.

Solution: (935)

SnO₂(s) + C(s) → CO2 + Sn

∆H = (∆f H)_P – (∆f H)_R

= – 394 + 581 = +187 KJ/mol^-1

∆S = (∆S)_P – (∆S)_R

= 210 + 52 – 56 – 6

= 200 J/k mol

∆G = 187 × 1000 – 200 × T

T = [187 × 1000] / 200 = 935 K

Question 3. In thermodynamics the P-V work done is given by w = −∫dVP_ext

For a system undergoing a particular process, the work done is, w = −∫dV(RT / V − b − a / V²)

This equation is applicable to a;

A. System that satisfies the van der Waals equation of state

B. Process that is reversible and isothermal

C. Process that is reversible and adiabatic

D. Process that is irreversible and at constant isothermal

Solution: (A, B and C)

w = −∫P_extdV

From van der Waal equation of state for one-mole gas.

(P + a / V² (v – b) = RT P = RT / V – b – a / V²

For reversible process

P_ext. = P_gas

W = −∫(RT/ V – b) – (a / v²)dv

So, the process is not applicable only for irreversible processes.

Hence, the correct options are a, b, c

Question 4. A reversible cyclic process for an ideal gas is shown below. Here, P, V and T are pressure, volume and temperature, respectively. The thermodynamic parameters q, w, H and U are heat, work, enthalpy and internal energy, respectively.

The correct option(s) is (are)

A. q_AC = ΔU_BC and w_AB = P₂(V₂ – V₁)

B. w_BC = P₂(V₂ – V₁) and q_BC = ΔH_AC

C. ΔH_CA < ΔU_CA and q_AC = ΔU_BC

D. q_BC = ΔH_AC and ΔH_CA > ΔU_CA

Solution: (B and C)

AC → Isochoric

AB → Isothermal

BC → Isobaric

# q_AC = ΔU_BC = nC_V (T₂ – T₁)

W_AB = nRT₁ ln (V₂/V₁) A (wrong)

# q_BC = ΔH_AC = nC_P (T₂ – T₁)

W_BC = – P₂(V₁ – V₂) B (correct)

# nC_P (T₁ – T₂) < nC_V(T₁ – T₂) C (correct)

ΔH_CA < ΔU_CA

# D (wrong)

Question 5. An ideal gas is expanded from (p₁, V₁, T₁) to (p₂, V₂, T₂)under different conditions. The correct statement(s) among the following is/are:

A. The work done on the gas is maximum when it is compressed irreversibly from (p₂, V₂) to (p₁, V₁) against constant pressure p₁

B. The work done by the gas is less when it is expanded reversibly from V₁ to V₂ under adiabatic conditions as compared to that when expanded reversibly from V₁ to V₂ under isothermal conditions

C. If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic

D. The change in internal energy of the gas is (i) zero if it is expanded reversibly with T₁ = T₂, and (ii) positive, if it is expanded reversibly under adiabatic conditions with T₁ ≠ T₂

Solution: (A, B and C)

(A) ω = −Pext ΔV

Work done on gas when compressed

ω = is positive

∴ equation becomes ω = Pext ΔV

(B) under adiabatic conditions

ω = (P₂V₂ − P₁V₁) / (γ − 1)

since equation V₂ > V₁

∴ ω = +ve (work done on the system)

for reversible condition (Isothermal)

ω = −nRTln V₂ / V₁

= V₂ > V₁

ω = −ve (work done on the system)

ω = −ve (for adiabatic reversibly work done by gas)

ω = +ve (for isothermal reversibly work done by gas)

(C) As free expansion in there (irreversible)

∴Pext=0

ω = (−Pext × ΔV) = 0 (Isothermal irreversible process)

In adiabatic ΔU = −ve

ΔU = q + ω

− = 0 + ω

ω = −negative

similarly, ΔG = ΔH − TΔs

gas expand ∴ Δs = +ve

ΔG = ΔH − Ve

= ΔV + pΔV − ve

= q + w + pΔc − ve

= o + w + w − ve

= −ve + (−ve) − ve

ΔG = −ve

∴ The process is spontaneous

∴ both occur simultaneously

Question 6. An ideal gas in a thermally insulated vessel at internal pressure = P₁, volume = V₁ and absolute temperature = T₁ expands irreversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute temperature of the gas are P₂, V₂ and T₂, respectively. For this expansion,

A. q = 0

B. T₂ = T₁

C. P₂ V₂ = P₁ V₁

D. P₂ V₁^γ = P₁V₁^γ

Solution: (C)

Since the container is thermally insulated q = 0, and it is a case of free expansion therefore W = 0 and ∆ = E 0

So, T₁ = T₂

Also, P₁V₁ = P₂V₂

Question 7 and 8 Paragraph: A fixed mass ‘m’ of gas is subjected to the transformation of the state from K to L to M and back to K as shown in the figure.

Question 7. The succeeding operations that enable this transformation of state are;

A. Heating, cooling, heating, cooling

B. Cooling, heating, cooling, heating

C. Heating, cooling, cooling, heating

D. Cooling, heating, heating, cooling

Solution: (C)

To reiterate the operations follow the path K, L. M, N. We simply need to use the ideal gas law here to get an idea of what happens. The ideal gas law states that:

PV = nRT

Since there is only a transformation of states number of moles (n) remains constant. Looking at these processes, in each of the steps either the pressure of the volume is constant. So using this law it is pretty easy to figure out what happens.

Clearly – K to L would be heating since volume is increasing (P is constant). L to M is interesting since P decreases keeping V constant. Now, this is possible only if cooling happens (i.e. T falls). This is a direct consequence of the ideal gas law.

Similarly, in steps M – N cooling happens : P is constant, V decreases which mean that T must fall. And lastly, N – K heating has to happen since V is constant, and P increases.

The process taking place at:

Constant temperature – isothermal

Constant pressure – isobaric

Constant volume – isochoric

Constant heat – adiabatic

Question 8. The pair of isochoric processes among the transformation of states is

A. K to L and L to M

B. L to M and N to K

C. L to M and M to N

D. M to N and N to K

Solution: (B)

Here you just need to know that isochoric means a constant volume process. Since V is on the X-axis, any process which is represented by a line perpendicular to the X-axis would be an isochoric one.

K-L heating, isobaric

L-M cooling, isochoric

M-N cooling, isobaric

N-K heating, isochoric

Question 9. The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are Δ_fG^∘C(graphite) = 0 kJ mol⁻¹

Δ_fG^∘C(diamond) = 2.9 kJ mol⁻¹

The standard state means that the pressure should be 1 bar, and the substance should be pure at a given temperature. The conversion of graphite C(graphite) to diamond C(diamond) reduces its volume by 2 × 10⁻⁶m³mol⁻¹. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is?

[Useful information: 1 J = 1 kgm²s⁻²; 1 Pa = 1 kg m⁻¹s⁻²; 1 bar = 105 Pa]

A. 14501 bar

B. 29001 bar

C. 1450 bar

D. 58001 bar

Solution: (A)

Cgraphite → Cdiamand

ΔG_r^o = ΔG_diamond^o − ΔG_graphite^o

2.9 KJmol⁻¹ − 0 KJmol⁻¹ = 2.9 KJmol⁻¹

ΔG_r^o = ΔH_r^o −TΔS_r^o

ΔS_r^o ≈ 0

TΔS_r^o ≈ 0

ΔH_r^o = ΔE_r^o + Δ(PV)

Δ_r^o= 0 because of isothermal process ΔT = 0

ΔH_r^o = 2.9 × 10³ = Δ(PV)

At equilibrium pressure is constant

2.9 × 10³ Jmol⁻¹ = P(V₂ − V₁)

P = 2.9 × 103 Jmol⁻¹ / 2 × 10⁻⁶ m³ mol⁻¹

= 1450 × 10⁶ pascal

= 14500 bar

Total pressure = equilibrium pressure + initial pressure = 14501bar

Hence, the correct option is A.

Question 10. For the process –

H2O(l) → H2O(g) at T = 100°C and 1 atmosphere,

The correct choice is:

(A) ∆S_system > 0 and ∆S_surrounding > 0

(B) ∆S_system > 0 and ∆S_surrounding < 0

(C) ∆S_system < 0 and ∆S_surrounding > 0

(D) ∆S_system < 0 and ∆S_surrounding < 0

Solution: (b)

At 100^∘C and 1-atmosphere pressure H₂O(l) ⇌ H₂O(g) is at equilibrium.

For equilibrium,

ΔS_total = 0 and ΔS_system + ΔS_surrounding = 0

∴ΔS_system>0 and ΔS_surrounding<0

Hence, the correct option is B.

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