A gas pushes a piston. Work is just force × distance , but force isn't constant — it depends on pressure, which changes as the gas expands. So we cannot multiply once; we must add up tiny pushes over the whole journey. That summation is an integral. Every process formula below is the same idea (W = ∫ P d V W=\int P\,dV W = ∫ P d V different path .
Definition Work done by a gasWhen a gas expands, it does positive work on surroundings; when compressed, work done by it is negative .
WHY start from force? Because work is fundamentally d W = F ⃗ ⋅ d x ⃗ dW = \vec F\cdot d\vec x d W = F ⋅ d x
HOW to get pressure in:
Piston of area A A A P P P F = P A F = PA F = P A
Piston moves out by d x dx d x d W = F d x = P A d x dW = F\,dx = PA\,dx d W = F d x = P A d x
But A d x = d V A\,dx = dV A d x = d V
d W = P d V ⟹ W = ∫ V 1 V 2 P d V \boxed{dW = P\,dV} \quad\Longrightarrow\quad W = \int_{V_1}^{V_2} P\,dV d W = P d V ⟹ W = ∫ V 1 V 2 P d V
This is the 80/20 core : master W = ∫ P d V W=\int P\,dV W = ∫ P d V P P P V V V
WHAT: P P P
W = ∫ V 1 V 2 P d V = P ∫ V 1 V 2 d V = = = P ( V 2 − V 1 ) = = W = \int_{V_1}^{V_2} P\,dV = P\int_{V_1}^{V_2} dV = ==P(V_2 - V_1)== W = ∫ V 1 V 2 P d V = P ∫ V 1 V 2 d V === P ( V 2 − V 1 ) ==
Using P V = n R T PV=nRT P V = n R T W = n R ( T 2 − T 1 ) W = nR(T_2 - T_1) W = n R ( T 2 − T 1 )
Worked example 2 mol gas heated 300 K → 400 K at constant pressureWhy this step? Isobaric, so use W = n R Δ T W=nR\Delta T W = n R Δ T W = 2 × 8.314 × ( 400 − 300 ) = 1662.8 J W = 2 \times 8.314 \times (400-300) = 1662.8\ \text{J} W = 2 × 8.314 × ( 400 − 300 ) = 1662.8 J
WHAT: V V V ⇒ d V = 0 \Rightarrow dV = 0 ⇒ d V = 0
W = ∫ V 1 V 1 P d V = = = 0 = = W = \int_{V_1}^{V_1} P\,dV = ==0== W = ∫ V 1 V 1 P d V === 0 ==
No volume change means the piston never moves — no distance, no work. All heat goes into internal energy.
WHY harder: P P P V V V P ( V ) P(V) P ( V )
HOW: Ideal gas at constant T T T P V = n R T ⇒ P = n R T V PV = nRT \Rightarrow P = \dfrac{nRT}{V} P V = n R T ⇒ P = V n R T
W = ∫ V 1 V 2 n R T V d V = n R T ∫ V 1 V 2 d V V = n R T [ ln V ] V 1 V 2 W = \int_{V_1}^{V_2} \frac{nRT}{V}\,dV = nRT\int_{V_1}^{V_2}\frac{dV}{V} = nRT\,[\ln V]_{V_1}^{V_2} W = ∫ V 1 V 2 V n R T d V = n R T ∫ V 1 V 2 V d V = n R T [ ln V ] V 1 V 2
W = = = n R T ln V 2 V 1 = = = n R T ln P 1 P 2 \boxed{W = ==nRT\ln\!\frac{V_2}{V_1}== = nRT\ln\!\frac{P_1}{P_2}} W === n R T ln V 1 V 2 === n R T ln P 2 P 1
(Used V 2 V 1 = P 1 P 2 \dfrac{V_2}{V_1}=\dfrac{P_1}{P_2} V 1 V 2 = P 2 P 1
Worked example 1 mol expands isothermally at 300 K, volume doublesWhy this step? Isothermal → log formula, V 2 / V 1 = 2 V_2/V_1=2 V 2 / V 1 = 2 W = 1 × 8.314 × 300 × ln 2 = 1728.8 J W = 1\times8.314\times300\times\ln 2 = 1728.8\ \text{J} W = 1 × 8.314 × 300 × ln 2 = 1728.8 J
WHY a new formula: here P V γ = K PV^\gamma = K P V γ = K P = K V − γ P = K V^{-\gamma} P = K V − γ
HOW:
W = ∫ V 1 V 2 K V − γ d V = K [ V − γ + 1 1 − γ ] V 1 V 2 = K 1 − γ ( V 2 1 − γ − V 1 1 − γ ) W = \int_{V_1}^{V_2} K V^{-\gamma}\,dV = K\left[\frac{V^{-\gamma+1}}{1-\gamma}\right]_{V_1}^{V_2} = \frac{K}{1-\gamma}\left(V_2^{1-\gamma}-V_1^{1-\gamma}\right) W = ∫ V 1 V 2 K V − γ d V = K [ 1 − γ V − γ + 1 ] V 1 V 2 = 1 − γ K ( V 2 1 − γ − V 1 1 − γ )
Now K = P 1 V 1 γ = P 2 V 2 γ K = P_1V_1^\gamma = P_2V_2^\gamma K = P 1 V 1 γ = P 2 V 2 γ K V 2 1 − γ = P 2 V 2 KV_2^{1-\gamma}=P_2V_2 K V 2 1 − γ = P 2 V 2 K V 1 1 − γ = P 1 V 1 KV_1^{1-\gamma}=P_1V_1 K V 1 1 − γ = P 1 V 1
W = P 1 V 1 − P 2 V 2 γ − 1 = = = n R ( T 1 − T 2 ) γ − 1 = = \boxed{W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = ==\frac{nR(T_1 - T_2)}{\gamma - 1}==} W = γ − 1 P 1 V 1 − P 2 V 2 === γ − 1 n R ( T 1 − T 2 ) ==
In adiabatic expansion, the work comes entirely from internal energy (W = − Δ U W = -\Delta U W = − Δ U cools (T 2 < T 1 T_2<T_1 T 2 < T 1 W > 0 W>0 W > 0
Worked example 1 mol monatomic gas (γ = 5 / 3 \gamma=5/3 γ = 5/3 W = 1 × 8.314 × ( 400 − 300 ) 5 / 3 − 1 = 831.4 0.667 = 1247 J W = \dfrac{1\times 8.314 \times (400-300)}{5/3 - 1} = \dfrac{831.4}{0.667} = 1247\ \text{J} W = 5/3 − 1 1 × 8.314 × ( 400 − 300 ) = 0.667 831.4 = 1247 J
For a closed loop, W n e t = ∮ P d V = W_{net} = \oint P\,dV = W n e t = ∮ P d V = area enclosed by the loop .
Clockwise loop → W > 0 W>0 W > 0 W < 0 W<0 W < 0
Common mistake Steel-manned errors(a) "Isothermal work is zero because Δ U = 0 \Delta U=0 Δ U = 0 Feels right because Δ T = 0 \Delta T=0 Δ T = 0 Δ U = 0 \Delta U=0 Δ U = 0 Fix: Δ U = 0 \Delta U=0 Δ U = 0 Q = W Q=W Q = W W = 0 W=0 W = 0 W = n R T ln ( V 2 / V 1 ) W=nRT\ln(V_2/V_1) W = n R T ln ( V 2 / V 1 ) (b) Using P Δ V P\Delta V P Δ V Feels right — it worked for isobaric. Fix: P P P n R T V \frac{nRT}{V} V n R T log .
(c) Sign of adiabatic work. Feels right to write P 2 V 2 − P 1 V 1 γ − 1 \frac{P_2V_2-P_1V_1}{\gamma-1} γ − 1 P 2 V 2 − P 1 V 1 Fix: It's P 1 V 1 − P 2 V 2 γ − 1 \frac{P_1V_1-P_2V_2}{\gamma-1} γ − 1 P 1 V 1 − P 2 V 2 W > 0 W>0 W > 0
Recall Feynman: explain to a 12-year-oldImagine pushing a swing. If you push with the same strength the whole way, work is just strength × distance (isobaric). But a gas gets weaker as it spreads out (pressure drops), so you have to add up lots of little pushes that keep shrinking — that's the integral. If the gas is wrapped in a blanket so no heat escapes (adiabatic), it uses up its own warmth to push, so it gets colder. If you keep it at the same warmth (isothermal), it can push longer and the total comes out as a "log."
"I Buy Very Cheap Apples" → I sothermal (n R T ln nRT\ln n R T ln B arely-moving = isoB aric (P Δ V P\Delta V P Δ V V olume-fixed = isochoric (0 0 0 A diabatic (n R Δ T γ − 1 \frac{nR\Delta T}{\gamma-1} γ − 1 n R Δ T remember the integral difficulty : Isochoric(0) < Isobaric(easy) < Isothermal(log) < Adiabatic(power) .
Starting point for all work calculations d W = P d V dW = P\,dV d W = P d V , so
W = ∫ V 1 V 2 P d V W=\int_{V_1}^{V_2}P\,dV W = ∫ V 1 V 2 P d V (area under P–V curve)
Work in isobaric process W = P ( V 2 − V 1 ) = n R Δ T W = P(V_2-V_1) = nR\Delta T W = P ( V 2 − V 1 ) = n R Δ T Work in isochoric process W = 0 W=0 W = 0 (since
d V = 0 dV=0 d V = 0 )
Work in isothermal process W = n R T ln ( V 2 / V 1 ) = n R T ln ( P 1 / P 2 ) W = nRT\ln(V_2/V_1) = nRT\ln(P_1/P_2) W = n R T ln ( V 2 / V 1 ) = n R T ln ( P 1 / P 2 ) Why isothermal gives a log P = n R T / V P=nRT/V P = n R T / V varies, and
∫ d V / V = ln V \int dV/V = \ln V ∫ d V / V = ln V Work in adiabatic process W = P 1 V 1 − P 2 V 2 γ − 1 = n R ( T 1 − T 2 ) γ − 1 W = \dfrac{P_1V_1-P_2V_2}{\gamma-1} = \dfrac{nR(T_1-T_2)}{\gamma-1} W = γ − 1 P 1 V 1 − P 2 V 2 = γ − 1 n R ( T 1 − T 2 ) In adiabatic expansion, where does work come from From internal energy, so gas cools (
W = − Δ U W=-\Delta U W = − Δ U )
Net work in a cyclic process Area enclosed by the loop on P–V diagram (clockwise = positive)
Is work a state function No, it is path-dependent (depends on the curve, not just endpoints)
First Law of Thermodynamics — Δ U = Q − W \Delta U = Q - W Δ U = Q − W W W W P-V Diagrams — work = area under curveAdiabatic Process and PV^gamma — source of P V γ = K PV^\gamma=K P V γ = K Isothermal Process and Boyle's Law Internal Energy and Cv — links adiabatic W = − Δ U W=-\Delta U W = − Δ U Carnot Cycle — net cyclic work
Master formula W = integral P dV
Isothermal W = nRT ln V2 over V1
Adiabatic W = nR T1-T2 over gamma-1
All heat to internal energy
Intuition Hinglish mein samjhoDekho, saara kaam (work) ek hi master formula se nikalta hai: d W = P d V dW = P\,dV d W = P d V F = P A F=PA F = P A d x dx d x A d x = d V A\,dx = dV A d x = d V P d V P\,dV P d V ∫ P d V \int P\,dV ∫ P d V P P P V V V area . Isi liye work path-dependent hai: same do points ke beech alag-alag raaste, alag area, alag work.
Ab bas yeh dekhna hai ki har process mein P P P Isobaric mein P P P W = P Δ V = n R Δ T W=P\Delta V = nR\Delta T W = P Δ V = n R Δ T Isochoric mein d V = 0 dV=0 d V = 0 W = 0 W=0 W = 0 Isothermal mein P = n R T / V P=nRT/V P = n R T / V ∫ d V / V = ln \int dV/V=\ln ∫ d V / V = ln W = n R T ln ( V 2 / V 1 ) W=nRT\ln(V_2/V_1) W = n R T ln ( V 2 / V 1 ) Adiabatic mein P V γ = K PV^\gamma=K P V γ = K W = P 1 V 1 − P 2 V 2 γ − 1 = n R ( T 1 − T 2 ) γ − 1 W=\frac{P_1V_1-P_2V_2}{\gamma-1}=\frac{nR(T_1-T_2)}{\gamma-1} W = γ − 1 P 1 V 1 − P 2 V 2 = γ − 1 n R ( T 1 − T 2 )
Important intuition: adiabatic expansion mein heat bahar nahi aata, toh gas apni hi internal energy kharch karke kaam karti hai — isliye gas thandi ho jaati hai (W = − Δ U W=-\Delta U W = − Δ U P Δ V P\Delta V P Δ V P P P
Ek common galti: "isothermal mein Δ U = 0 \Delta U=0 Δ U = 0 W = 0 W=0 W = 0 Δ U = 0 \Delta U=0 Δ U = 0 Q = W Q=W Q = W
