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Thermodynamics

The Thermodynamics category gathers questions that cover the fundamental principles of heat, energy, work, and the laws governing them. It helps learners review concepts such as temperature, entropy, and energy transfer, providing clear explanations and practice problems.

268 questions

What distinguishes work from heat as a mode of energy transfer?

**Energy transfer** first law ΔU = Q - W, W includes P-V work, shaft work, electrical work, Q includes conduction Fourier law, convection, radiation Stefan-Boltzmann, distinction important because work is controllable, heat spontaneous from hot to cold, entropy associated with heat not work, explaining why heat engine efficiency

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

A solid of 2 moles is heated from 300 K to 320 K . If its molar specific heat capacity is 25.5 J mol⁻¹ K⁻¹ , what is the

**Work and heat** both energy transfer modes, work organized, e.g., lifting weight, compressing gas, electrical current, heat random due to temperature difference, work can be completely converted to heat via friction, but heat cannot be completely converted to work (second law), energy transfer modes include work (mechanical, electrical) and heat (conduction, convection, radiation). Δ Q = μ C Δ T . μ = 2 , C = 25.5 , Δ T = 320 - 300 = 20 . Δ Q = 2 × 25.5 × 20 = 1020 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

In an isobaric process, 1.2 moles of gas expand from 350 K to 420 K . What is the heat supplied if C_p = 25.5 J mol⁻¹ K⁻

**Modes of energy transfer** work is force times displacement, e.g., gas expansion W = P ΔV, heat is due to temperature gradient, internal energy change same for different combinations of Q and W, e.g., same ΔU can be achieved by adding heat at constant volume or doing work adiabatically, illustrating equivalence but distinction in mechanism. Δ Q = μ C_p Δ T . μ = 1.2 , C_p = 25.5 , Δ T = 420 - 350 = 70 . Δ Q = 1.2 × 25.5 × 70 = 2142 J . Using first law ΔU = Q - W, W = ∫ P dV,

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

A gas expands adiabatically, doing 360 J of work. What is the change in its internal energy?

**Energy transfer** first law ΔU = Q - W, W includes P-V work, shaft work, electrical work, Q includes conduction Fourier law, convection, radiation Stefan-Boltzmann, distinction important because work is controllable, heat spontaneous from hot to cold, entropy associated with heat not work, explaining why heat engine efficiency

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

A system in a cyclic process absorbs 940 J of heat and rejects 360 J . What is the net work done?

**Work and heat** both energy transfer modes, work organized, e.g., lifting weight, compressing gas, electrical current, heat random due to temperature difference, work can be completely converted to heat via friction, but heat cannot be completely converted to work (second law), energy transfer modes include work (mechanical, electrical) and heat (conduction, convection, radiation). For cyclic: Δ U = 0 , Q_net = W . Q_net = Q_absorb - Q_reject = 940 - 360 = 580 J . W = 580 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

Which of the following statements is correct regarding the Zeroth Law of Thermodynamics?

**Modes of energy transfer** work is force times displacement, e.g., gas expansion W = P ΔV, heat is due to temperature gradient, internal energy change same for different combinations of Q and W, e.g., same ΔU can be achieved by adding heat at constant volume or doing work adiabatically, illustrating equivalence but distinction in mechanism. The Zeroth Law states that if two systems are in thermal equilibrium with a third, they are in equilibrium with each other, establishing temperature as a measurable property. It does not involve heat flow direction or work, which are addressed by other laws. Using first law ΔU = Q -

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

A gas expands adiabatically from 9 atm and 18 L to 3 atm . What is the final volume? ( gamma = 1.4 )

**Energy transfer** first law ΔU = Q - W, W includes P-V work, shaft work, electrical work, Q includes conduction Fourier law, convection, radiation Stefan-Boltzmann, distinction important because work is controllable, heat spontaneous from hot to cold, entropy associated with heat not work, explaining why heat engine efficiency

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

How much heat is required to vaporize 0.9 g of water at 100^circ C and 1 atm ? (Latent heat = 2256 J/g )

**Work and heat** both energy transfer modes, work organized, e.g., lifting weight, compressing gas, electrical current, heat random due to temperature difference, work can be completely converted to heat via friction, but heat cannot be completely converted to work (second law), energy transfer modes include work (mechanical, electrical) and heat (conduction, convection, radiation). Δ Q = m L . m = 0.9 , L = 2256 . Δ Q = 0.9 × 2256 = 2030.4 J ≈ 2030 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁),

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

Which of the following statements is incorrect regarding the Second Law of Thermodynamics?

**Modes of energy transfer** work is force times displacement, e.g., gas expansion W = P ΔV, heat is due to temperature gradient, internal energy change same for different combinations of Q and W, e.g., same ΔU can be achieved by adding heat at constant volume or doing work adiabatically, illustrating equivalence but distinction in mechanism. The Second Law limits efficiency and directionality (e.g., no 100% heat-to-work conversion, Kelvin-Planck). Option B is incorrect; it contradicts the law, which requires heat rejection to a cold reservoir. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W =

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

A monatomic gas undergoes an adiabatic expansion from 700 K to 350 K with 2 moles . What is the work done? ( R = 8.3 J m

**Energy transfer** first law ΔU = Q - W, W includes P-V work, shaft work, electrical work, Q includes conduction Fourier law, convection, radiation Stefan-Boltzmann, distinction important because work is controllable, heat spontaneous from hot to cold, entropy associated with heat not work, explaining why heat engine efficiency

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

In an adiabatic process, what compensates for the absence of heat transfer?

**Work and heat** both energy transfer modes, work organized, e.g., lifting weight, compressing gas, electrical current, heat random due to temperature difference, work can be completely converted to heat via friction, but heat cannot be completely converted to work (second law), energy transfer modes include work (mechanical, electrical) and heat (conduction, convection, radiation). In an adiabatic process ( Δ Q = 0 ), the change in internal energy ( Δ U ) is entirely due to work done ( Δ U = -Δ W ). Work done by or on the system adjusts the internal energy, as no heat is exchanged. Using first law ΔU

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

A gas at 10 atm and 80^circ C in a 8 L container is heated isochorically to 140^circ C . What is the final pressure?

**Modes of energy transfer** work is force times displacement, e.g., gas expansion W = P ΔV, heat is due to temperature gradient, internal energy change same for different combinations of Q and W, e.g., same ΔU can be achieved by adding heat at constant volume or doing work adiabatically, illustrating equivalence but distinction in mechanism. For isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 10 atm , T₁ = 80 + 273 = 353 K , T₂ = 140 + 273 = 413 K . (10)/(353) = (P₂)/(413) ⇒ P₂ = (10 × 413)/(353) ≈ 11.7 atm . Using first law ΔU = Q - W,

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes