Welcome to our complete and easy-to-understand Class 11 Chemistry Chapter 5 Thermodynamics NCERT Solutions. In this chapter, you will explore the fundamental concepts of thermodynamics, including system and surroundings, types of processes, internal energy, work, heat, enthalpy, and the laws of thermodynamics.

Class 11 Chemistry Chapter 5 Thermodynamics NCERT Solutions

All NCERT in-text questions and numerical problems are solved in plain text format, making it simple for CBSE students to grasp key ideas and perform well in exams. These Class 11 Chemistry Chapter 5 Thermodynamics NCERT Solutions follow the latest CBSE syllabus and are perfect for quick revision and deep understanding.

Class 11 Chemistry Chapter 5 Thermodynamics NCERT Solutions

Problem 5.1

Q: Express the change in internal energy of a system when:
(i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system. What type of wall does the system have?
(ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. What type of wall does the system have?
(iii) w amount of work is done by the system and q amount of heat is supplied to the system. What type of system would it be?

A:
(i) ΔU = w, wall is adiabatic
(ii) ΔU = –q, thermally conducting walls
(iii) ΔU = q – w, closed system

Problem 5.2

Q: Two litres of an ideal gas at a pressure of 10 atm expands isothermally at 25°C into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion?

A:
q = –w = pₑₓ × (10 – 2) = 0 × 8 = 0
So, no work is done; no heat is absorbed.

Problem 5.3

Q: Consider the same expansion as in problem 5.2, but this time against a constant external pressure of 1 atm.

A:
q = –w = pₑₓ × (10 – 2) = 1 × 8 = 8 L·atm

Problem 5.4

Q: Consider the expansion given in problem 5.2 for 1 mol of an ideal gas conducted reversibly.

A:
w = –2.303 × nRT × log(Vf / Vi)
= –2.303 × 1 × 0.08206 × 298 × log(10 / 2)
= –2.303 × 0.08206 × 298 × 0.6990
= –39.366 L·atm
q = –w = +39.366 L·atm

Problem 5.5

Q: If water vapour is assumed to be a perfect gas, molar enthalpy change for vaporisation of 1 mol of water at 1 bar and 100°C is 41 kJ mol⁻¹. Calculate the internal energy change.

A:
ΔU = ΔH – Δn_gas × R × T
= 41.00 – (1 × 8.314 × 373 × 10⁻³)
= 41.00 – 3.096
= 37.904 kJ mol⁻¹

Problem 5.6

Q: 1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atm. The temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7 kJ/K, what is the enthalpy change for the reaction?
C(graphite) + O₂(g) → CO₂(g)

A:
q = –CV × ΔT = –20.7 × 1 = –20.7 kJ
ΔU for 1 g graphite = –20.7 kJ
Molar enthalpy = –20.7 × (12 / 1) = –248.4 kJ mol⁻¹
ΔH = ΔU = –248.4 kJ mol⁻¹ (since Δn_gas = 0)

Problem 5.7

Q: A swimmer comes out from a pool covered with a film of 18 g of water. How much heat must be supplied to evaporate this water at 298 K? Also calculate the internal energy of vaporisation.
Given: ΔH_vap = 44.01 kJ/mol

A:
Moles of water = 18 / 18 = 1 mol
q = 1 × 44.01 = 44.01 kJ
ΔU = 44.01 – (1 × 8.314 × 298 × 10⁻³)
= 44.01 – 2.48 = 41.53 kJ

Problem 5.8

Q: Calculate the internal energy change when 1 mol of water at 100°C and 1 bar is converted to ice at 0°C.
Given:
ΔH_fusion = 6.00 kJ/mol
Specific heat of water = 4.2 J/g°C

A:
Step 1: Cooling water from 100°C to 0°C:
ΔH₁ = –(18 × 4.2 × 100) J = –7560 J = –7.56 kJ

Step 2: Freezing water at 0°C:
ΔH₂ = –6.00 kJ

Total enthalpy change:
ΔH = ΔH₁ + ΔH₂ = –7.56 – 6.00 = –13.56 kJ
ΔU = ΔH = –13.56 kJ

Problem 5.9

Q: The combustion of 1 mol of benzene at 298 K and 1 atm liberates 3267.0 kJ of heat. Calculate the standard enthalpy of formation (ΔfH⁰) of benzene.
Given:
ΔfH⁰[CO₂(g)] = –393.5 kJ/mol
ΔfH⁰[H₂O(l)] = –285.83 kJ/mol

A:
ΔfH⁰ = [6 × (–393.5) + 3 × (–285.83)] – (–3267.0)
= –3218.49 + 3267.0 = +48.51 kJ/mol

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In this page, we have provided detailed solutions for all the important questions from Class 11 Chemistry Chapter 5 Thermodynamics NCERT, including numerical and conceptual problems. These answers will help students build a strong understanding of core thermodynamic concepts like internal energy, enthalpy, work, heat, and state functions.

The Class 11 Chemistry Chapter 5 Thermodynamics NCERT solutions are presented in simple language and plain text format, making them ideal for quick revision before exams. Every answer aligns with the latest CBSE guidelines and textbook requirements.

Whether you are preparing for class tests or final board exams, this Class 11 Chemistry Chapter 5 Thermodynamics NCERT content will support your learning and improve your problem-solving skills.