Which of the following statements is incorrect about an isochoric process?
**Adiabatic work** W = ∫ P dV = (P₁V₁ - P₂V₂)/(γ-1), for expansion V₂>V₁ P₂
Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination
Adiabatic Processes and Gamma Determination
This category explores adiabatic processes and the methods used to determine the heat‑capacity ratio γ. It covers theory, examples of adiabatic expansion and compression, and practical ways to calculate γ from data.
Filter questions
A gas at 7 atm and 40^circ C in a 5 L container is cooled isochorically to -20^circ C . What is the final pressure?
**Gamma determination** γ = C_p/C_v, C_p - C_v = R, for monatomic f=3 C_v=3/2 R C_p=5/2 R γ=1.67, diatomic f=5 C_v=5/2 R C_p=7/2 R γ=1.4, adiabatic relation P V^γ = const allows γ determination from P-V measurements, slope of log P vs log V = -γ. For isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 7 atm , T₁ = 40 + 273 = 313 K , T₂ = -20 + 273 = 253 K . (7)/(313) = (P₂)/(253) ⇒ P₂ = (7 × 253)/(313) ≈ 5.66 atm . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P
Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination
A system releases 300 J of heat and has 150 J of work done on it. What is the change in internal energy?
**Adiabatic process** no heat exchange Q=0, first law ΔU = -W, for ideal gas P V^γ = constant, T V^{γ-1}= constant, P^{1-γ} T^{γ}= constant, γ = C_p/C_v = (f+2)/f, monatomic γ=5/3, diatomic γ=7/5. Work done W = (P₁V₁ - P₂V₂)/(γ-1), temperature changes due to work. First Law: Δ Q = Δ U + Δ W . Δ Q = -300 J (heat released), Δ W = -150 J (work done on system, negative by convention). -300 = Δ U + (-150) . Δ U = -300 + 150 = -150 J . Using first law ΔU = Q - W, W = ∫ P dV,
Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination
A gas undergoes an adiabatic compression from 20 L to 5 L , increasing its pressure from 2 atm to 16 atm . What is the v
**Adiabatic work** W = ∫ P dV = (P₁V₁ - P₂V₂)/(γ-1), for expansion V₂>V₁ P₂
Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination
What is the significance of a quasi-static process in thermodynamics?
**Gamma determination** γ = C_p/C_v, C_p - C_v = R, for monatomic f=3 C_v=3/2 R C_p=5/2 R γ=1.67, diatomic f=5 C_v=5/2 R C_p=7/2 R γ=1.4, adiabatic relation P V^γ = const allows γ determination from P-V measurements, slope of log P vs log V = -γ. A quasi-static process is infinitely slow, ensuring the system remains in thermal and mechanical equilibrium with its surroundings at every stage. This allows well-defined state variables (e.g., P , T ) and is an idealized condition for reversible processes. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W
Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination
Why does an isobaric process involve both internal energy change and work?
**Adiabatic process** no heat exchange Q=0, first law ΔU = -W, for ideal gas P V^γ = constant, T V^{γ-1}= constant, P^{1-γ} T^{γ}= constant, γ = C_p/C_v = (f+2)/f, monatomic γ=5/3, diatomic γ=7/5. Work done W = (P₁V₁ - P₂V₂)/(γ-1), temperature changes due to work. In an isobaric process ( P = constant ), heat added ( Δ Q ) increases internal energy ( Δ U ) and does work ( W = P Δ V ) due to volume expansion, as per the First Law: Δ Q = Δ U + P Δ V . Using first law ΔU = Q - W, W
Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination
How much work is done by 1 mole of an ideal gas in an isothermal expansion from 3 L to 9 L at 500 K ? ( R = 8.3 J mol⁻¹
**Adiabatic work** W = ∫ P dV = (P₁V₁ - P₂V₂)/(γ-1), for expansion V₂>V₁ P₂
Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination
A system absorbs 710 J of heat and performs 260 J of work. What is the change in internal energy?
**Gamma determination** γ = C_p/C_v, C_p - C_v = R, for monatomic f=3 C_v=3/2 R C_p=5/2 R γ=1.67, diatomic f=5 C_v=5/2 R C_p=7/2 R γ=1.4, adiabatic relation P V^γ = const allows γ determination from P-V measurements, slope of log P vs log V = -γ. First Law: Δ Q = Δ U + Δ W . Δ Q = 710 , Δ W = 260 (work by system). 710 = Δ U + 260 ⇒ Δ U = 710 - 260 = 450 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal
Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination
How much heat is required to raise the temperature of 0.2 kg of silver from 30^circ C to 50^circ C ? (Specific heat of s
**Adiabatic process** no heat exchange Q=0, first law ΔU = -W, for ideal gas P V^γ = constant, T V^{γ-1}= constant, P^{1-γ} T^{γ}= constant, γ = C_p/C_v = (f+2)/f, monatomic γ=5/3, diatomic γ=7/5. Work done W = (P₁V₁ - P₂V₂)/(γ-1), temperature changes due to work. Δ Q = m s Δ T . m = 0.2 , s = 236.1 , Δ T = 50 - 30 = 20 . Δ Q = 0.2 × 236.1 × 20 = 944.4 J ≈ 944 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W
Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination
A gas at 6 atm and 50^circ C in a 4 L container is heated isochorically to 150^circ C . What is the final pressure?
**Adiabatic work** W = ∫ P dV = (P₁V₁ - P₂V₂)/(γ-1), for expansion V₂>V₁ P₂
Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination
Which of the following statements is correct about isothermal processes for an ideal gas?
**Gamma determination** γ = C_p/C_v, C_p - C_v = R, for monatomic f=3 C_v=3/2 R C_p=5/2 R γ=1.67, diatomic f=5 C_v=5/2 R C_p=7/2 R γ=1.4, adiabatic relation P V^γ = const allows γ determination from P-V measurements, slope of log P vs log V = -γ. In an isothermal process ( T = constant ), Δ U = 0 for an ideal gas, and P V = constant . Option C is correct. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η
Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination
Which of the following statements is correct about the Second Law of Thermodynamics?
**Adiabatic process** no heat exchange Q=0, first law ΔU = -W, for ideal gas P V^γ = constant, T V^{γ-1}= constant, P^{1-γ} T^{γ}= constant, γ = C_p/C_v = (f+2)/f, monatomic γ=5/3, diatomic γ=7/5. Work done W = (P₁V₁ - P₂V₂)/(γ-1), temperature changes due to work. The Second Law (Clausius statement) states that heat cannot flow from a colder to a hotter body without work, reflecting natural directionality. Option C is correct. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1
Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination