First law for closed systems

Why it matters

The first law is energy bookkeeping: nothing is created or lost, it only moves across the boundary as heat or work or is stored inside the system. For a closed system — a charge in an engine cylinder, gas in a pressure vessel, an air spring — it turns a physical description into one equation you can solve for the unknown temperature, heat or work. Every cycle analysis later in the subject is built on it.

Key ideas

First law for a cycle. Joule's experiments showed that when a closed system executes a cycle, the net heat supplied equals the net work delivered: ∮δQ = ∮δW. No machine can produce net work over a cycle without an equal net heat input — a perpetual-motion machine of the first kind (PMM1) is impossible.

First law for a process. Since ∮(δQ − δW) = 0 for every cycle, the quantity (δQ − δW) is the change of a property — the total energy E. For a process 1→2:

• Q − W = ΔE = ΔU + ΔKE + ΔPE
• For a stationary closed system (the usual case) ΔKE = ΔPE = 0, so Q − W = ΔU.

Q and W are path functions; ΔU is a property change and depends only on the end states. That is why you can find ΔU along any convenient path.

Internal energy is the energy stored in the molecular structure — translational, rotational and vibrational kinetic energy plus intermolecular potential energy. For an ideal gas it depends on temperature only (Joule's free-expansion experiment), so du = cv·dT for any process of an ideal gas, not just constant-volume ones.

Enthalpy H = U + pV is a property, convenient whenever pressure is constant. For a closed system at constant pressure with only boundary work, Q = ΔH. For an ideal gas dh = cp·dT.

Specific heats. cv = (∂u/∂T)v, cp = (∂h/∂T)p. For an ideal gas cp − cv = R and γ = cp/cv. For air at room temperature cp ≈ 1.005 kJ/(kg·K), cv ≈ 0.718 kJ/(kg·K), γ ≈ 1.4. Solids and liquids are nearly incompressible, so cp ≈ cv = c.

Special cases.

• Constant volume: W = 0, so Q = ΔU = m·cv·ΔT.
• Constant pressure: W = p·ΔV, so Q = ΔH = m·cp·ΔT.
• Isothermal ideal gas: ΔU = 0, so Q = W.
• Adiabatic: Q = 0, so W = −ΔU (expansion cools the gas).
• Isolated system: Q = W = 0, so E is constant.
• Stirring (paddle-wheel) or electrical work is work input, W < 0 in the work-by-system convention.

Polytropic ideal-gas process pVⁿ = C. Combining W = mR(T₁ − T₂)/(n − 1) with ΔU gives Q = W·(γ − n)/(γ − 1). For 1 < n < γ, heat is added during expansion even though the gas cools.

Formulas

∮δQ = ∮δW (cycle) Q − W = ΔU + ΔKE + ΔPE (closed system, process) ΔU = m·cv·(T₂ − T₁) (ideal gas, any process) ΔH = m·cp·(T₂ − T₁) (ideal gas, any process) H = U + p·V cp − cv = R, γ = cp / cv Q = W·(γ − n) / (γ − 1) (ideal-gas polytropic process)

• Q: heat added to the system, kJ; W: work done by the system, kJ
• U, H: internal energy and enthalpy, kJ (u, h per kg: kJ/kg)
• m: mass, kg; T: absolute temperature, K
• cv, cp: specific heats at constant volume and pressure, kJ/(kg·K)
• R: specific gas constant, kJ/(kg·K); γ: ratio of specific heats; n: polytropic index

Worked examples

Example 1 (standard). A rigid tank holds 0.5 kg of air at 300 K. A paddle wheel does 20 kJ of work on the air while 5 kJ of heat leaks out. Find the final temperature. cv = 0.718 kJ/(kg·K).

1. Rigid tank: no boundary work. Paddle work is input, so W = −20 kJ. Heat leaves, so Q = −5 kJ.
2. ΔU = Q − W = −5 − (−20) = +15 kJ
3. ΔU = m·cv·(T₂ − T₁) → T₂ − T₁ = 15 / (0.5 × 0.718) = 41.78 K
4. T₂ = 300 + 41.78 = 341.8 K

Example 2 (GATE level). 0.2 kg of air at 300 K expands in a piston–cylinder following pV¹·³ = C until its volume triples. Find the work, the change in internal energy and the heat transfer. R = 0.287, cv = 0.718 kJ/(kg·K).

1. T₂ = T₁·(V₁/V₂)ⁿ⁻¹ = 300 × (1/3)⁰·³ = 215.77 K
2. W = m·R·(T₁ − T₂)/(n − 1) = 0.2 × 0.287 × 84.23 / 0.3 = 16.12 kJ
3. ΔU = m·cv·(T₂ − T₁) = 0.2 × 0.718 × (−84.23) = −12.10 kJ
4. Q = ΔU + W = −12.10 + 16.12 = +4.02 kJ (heat added)
5. Cross-check: Q = W(γ − n)/(γ − 1) = 16.12 × 0.1/0.4 = 4.03 kJ (taking γ = 1.4; the tiny difference is because 1.005/0.718 = 1.3997).

Answer: W = 16.1 kJ, ΔU = −12.1 kJ, Q = +4.0 kJ

Common mistakes

• Mixing sign conventions: if W is "work done on the system", the law is ΔU = Q + W.
• Using ΔU = m·cv·ΔT only for constant-volume processes. For an ideal gas it holds for every process.
• Using cv where cp is needed (constant-pressure heating), or vice versa.
• Treating paddle-wheel or electrical work as positive.
• Assuming the gas cools only when heat is removed — adiabatic expansion cools it too.
• Writing Q = ΔH for a process that is not at constant pressure.

For GATE AE

Expect numericals combining the first law with ideal-gas processes: rigid tanks with stirring or electrical work, constant-pressure heating with a piston, and polytropic expansion asking for Q given n and γ. Conceptual questions test PMM1, internal energy as a function of T for an ideal gas, and Q = ΔH at constant pressure. Practise fixing the sign of each interaction before writing numbers.

Quick check

1. A closed system receives 300 J of heat and does 100 J of work. Find ΔU.
2. In a cycle a system receives 50 kJ of heat. What is the net work?
3. For an ideal gas undergoing an isothermal process, what is ΔU?
4. Air at constant pressure is heated from 300 K to 400 K. Find the heat per kg (cp = 1.005 kJ/(kg·K)).
5. In an adiabatic expansion, does the temperature of an ideal gas rise or fall?

Answers: 1. 200 J. 2. 50 kJ. 3. Zero. 4. 100.5 kJ/kg. 5. It falls.