Zeroth & First Law: Energy Analysis

Thermodynamics is a heavy GATE ME subject (~10–14 marks), and the first law underpins every cycle in it. Closed-system and steady-flow energy balances are near-certain 1–2 mark items.

Zeroth law & definitions

Zeroth law: if two bodies are each in thermal equilibrium with a third, they are in equilibrium with each other — this defines temperature. System types: closed (no mass crossing), open/control volume (mass crosses), isolated (nothing crosses).

First law (energy conservation)

• Closed system: Q − W = ΔU (heat added minus work done = change in internal energy). For a cycle, ∮δQ = ∮δW.
• Enthalpy h = u + pv. Ideal gas: pv = RT; c_p − c_v = R; γ = c_p/c_v. Δu = c_v·ΔT, Δh = c_p·ΔT.
• Open system (SFEE, steady flow): q − w_s = Δh + ΔKE + ΔPE. For a turbine/compressor (adiabatic, ΔKE, ΔPE small): w_s = h₁ − h₂.

Process work (closed system)

• Isobaric (p const): W = p·ΔV. Isochoric (V const): W = 0.
• Isothermal (ideal gas): W = p₁V₁·ln(V₂/V₁).
• Adiabatic (pV^γ = const): W = (p₁V₁ − p₂V₂)/(γ − 1).
• Polytropic (pVⁿ = const): W = (p₁V₁ − p₂V₂)/(n − 1).

Exam Tricks & Tips

• 🎯 Closed system Q − W = ΔU; steady flow q − w_s = Δh — pick the right form: use ΔU for a piston, Δh for a turbine/nozzle.
• 🎯 c_p − c_v = R and γ = c_p/c_v — for air c_p ≈ 1.005, c_v ≈ 0.718 kJ/kgK, γ ≈ 1.4.
• 🎯 Δu = c_v ΔT and Δh = c_p ΔT hold for an ideal gas in ANY process (not just constant volume/pressure) — a common misconception.
• 🎯 Turbine/compressor work (adiabatic) = h₁ − h₂ — the SFEE workhorse; kinetic/potential terms usually negligible.
• 🎯 Adiabatic work = (p₁V₁ − p₂V₂)/(γ−1); polytropic uses (n−1) — same form, different exponent.
• ❌ Common mistake: applying the closed-system Q − W = ΔU to a flow device — turbines/compressors are open systems, so use the steady-flow energy equation with enthalpy.

Expected exam pattern

A 1-mark first-law, process-work, or c_p/c_v/γ NAT, and a 2-mark closed-system or SFEE problem (turbine/compressor work, nozzle exit velocity). The "Δu = c_vΔT always for ideal gas" fact and choosing ΔU vs Δh are common points.

Quick recap

Zeroth law defines temperature. Closed system Q − W = ΔU; cycle ∮δQ = ∮δW. Ideal gas pv = RT, c_p − c_v = R, γ = c_p/c_v; Δu = c_vΔT, Δh = c_pΔT (any process). SFEE q − w_s = Δh + ΔKE + ΔPE; turbine w = h₁ − h₂. Adiabatic W = (p₁V₁−p₂V₂)/(γ−1).

Cover the answer, recall, then check. 12 cards on the first law for GATE ME.

Q1. State the zeroth law of thermodynamics.
A1. If two systems are each in thermal equilibrium with a third, they are in equilibrium with each other — this defines temperature.

Q2. State the first law for a closed system.
A2. Q − W = ΔU (heat in minus work out equals the change in internal energy). For a cycle, ∮δQ = ∮δW.

Q3. Define enthalpy.
A3. h = u + pv (internal energy plus flow work). For an ideal gas, Δh = c_p·ΔT.

Q4. Relate c_p, c_v, R, and γ for an ideal gas.
A4. c_p − c_v = R and γ = c_p/c_v. For air, c_p ≈ 1.005, c_v ≈ 0.718 kJ/kgK, γ ≈ 1.4.

Q5. State the steady-flow energy equation.
A5. q − w_s = Δh + ΔKE + ΔPE, per unit mass, across a control volume.

Q6. Turbine/compressor work for adiabatic steady flow?
A6. w_s = h₁ − h₂ (kinetic and potential changes neglected).

Q7. Do Δu = c_vΔT and Δh = c_pΔT hold only for constant V/p?
A7. No — for an ideal gas they hold in ANY process, since u and h depend only on temperature.

Q8. Work done in an isothermal ideal-gas process?
A8. W = p₁V₁·ln(V₂/V₁) = mRT·ln(V₂/V₁).

Q9. Work in a reversible adiabatic (pV^γ = const) process?
A9. W = (p₁V₁ − p₂V₂)/(γ − 1).

Q10. Work in a polytropic process pVⁿ = const?
A10. W = (p₁V₁ − p₂V₂)/(n − 1).

Q11. Work in an isochoric (constant-volume) process?
A11. Zero — no volume change means no boundary work; all heat goes to internal energy.

Q12. Which law says a piston uses ΔU but a turbine uses Δh?
A12. The first law — closed systems (piston) use Q − W = ΔU; open systems (turbine) use the SFEE with enthalpy.

Thermodynamics is the accounting of energy, and the first law is its balance sheet: energy is conserved, it just changes form. GATE builds many numericals on closed-system and control-volume energy balances, so the sign conventions and the steady-flow energy equation must be automatic.

Core concept: the zeroth law defines temperature (bodies in thermal equilibrium share it); the first law states that energy is conserved — heat and work transfers change a system's internal energy.

Deep explanation

Beginner — the zeroth and first laws

• Zeroth law: if A is in thermal equilibrium with C and B is too, then A and B are in equilibrium — this is what makes temperature measurable.
• First law (closed system): Q − W = ΔU, where Q = heat added to the system, W = work done by the system, ΔU = change in internal energy. Sign convention (classical): heat in positive, work out positive.

Intermediate — properties and processes

• Internal energy U and enthalpy H = U + pV are state properties; heat and work are path functions (depend on the process).
• For an ideal gas: ΔU = m c_v ΔT, ΔH = m c_p ΔT, and c_p − c_v = R, γ = c_p/c_v.
• Common processes: isothermal (T const, W = mRT ln(V₂/V₁)), adiabatic (Q = 0, pV^γ = const), isobaric (W = pΔV), isochoric (V const, W = 0). For a cycle, ΔU = 0 so net Q = net W.

Advanced — the steady-flow energy equation (SFEE)

For an open system (control volume) at steady state, per unit mass:
q − w_s = (h₂ − h₁) + (V₂² − V₁²)/2 + g(z₂ − z₁),
where w_s is shaft work and h is specific enthalpy. Applying SFEE with the right terms dropped gives device relations:

• Nozzle (w_s = 0, adiabatic): h₁ + V₁²/2 = h₂ + V₂²/2 (enthalpy → kinetic energy).
• Turbine/compressor (Δke, Δpe ≈ 0): w_s = h₁ − h₂.
• Throttle valve: h₁ = h₂ (isenthalpic).
• Heat exchanger: q = h₂ − h₁ (w_s = 0).

Worked example

Steam enters a turbine at h₁ = 3200 kJ/kg and leaves at h₂ = 2400 kJ/kg. Inlet velocity 50 m/s, exit 120 m/s; adiabatic, neglect potential energy. Find the specific work output.
SFEE: w_s = (h₁ − h₂) + (V₁² − V₂²)/2.
Enthalpy term = 3200 − 2400 = 800 kJ/kg.
KE term = (50² − 120²)/2 = (2500 − 14,400)/2 = −5,950 J/kg = −5.95 kJ/kg.
w_s = 800 − 5.95 = 794 kJ/kg. (The kinetic-energy change is small — often neglected in turbines.)

GATE relevance

First-law closed-system and SFEE control-volume balances are among the most tested Thermodynamics skills, underpinning turbines, compressors, nozzles, throttles and all the power cycles. The device-specific SFEE reductions (turbine w = Δh, throttle h = const) are essential.

Exam tricks & shortcuts

• Fix your sign convention (Q in +, W out +) and keep it — mixing conventions flips answers.
• Throttling is isenthalpic (h₁ = h₂); a turbine gives w = h₁ − h₂; a nozzle converts Δh to Δke.
• For any cycle, ΔU = 0, so net heat = net work.
• Mnemonic: "Q minus W equals delta-U."

Key formulas

• First law (closed system): Q − W = ΔU; per cycle ΣQ = ΣW.
• Open system (SFEE): Q − W_s = ṁ[(h₂−h₁) + (V₂²−V₁²)/2 + g(z₂−z₁)].
• Work: W = ∫P dV; enthalpy h = u + Pv.
• Zeroth law: thermal equilibrium is transitive → defines temperature.

The first law is energy conservation; the steady-flow energy equation applies to turbines, pumps and nozzles.

Worked Example

Problem: A gas in a piston-cylinder device expands at constant pressure 200 kPa from a volume of 0.1 m³ to 0.3 m³. During the process 100 kJ of heat is added to the gas. Find the boundary work done and the change in internal energy of the gas.

Solution:
Boundary (displacement) work at constant pressure:
W = P·ΔV = P·(V₂ − V₁) = 200 kPa × (0.3 − 0.1) m³ = 200 × 0.2 = 40 kJ.
(Positive because the gas expands and does work on the surroundings.)

Apply the first law of thermodynamics for a closed system:
Q = ΔU + W ⇒ ΔU = Q − W.
ΔU = 100 − 40 = 60 kJ.

So of the 100 kJ of heat supplied, 40 kJ leaves as work and 60 kJ is stored as internal energy (raising the temperature).

Answer: Boundary work W = 40 kJ and internal energy change ΔU = +60 kJ.