Intuition The big pictureThe First Law is just conservation of energy go somewhere . It can either (1) raise the gas's internal energy
(make molecules jiggle faster → hotter), or (2) be spent by the gas pushing a
piston outward (doing work). Nothing else. That's the whole law.
Definition The three quantities
U U U internal energy : total microscopic energy (kinetic + potential of all molecules). A state function — depends only on the current state ( P , V , T ) (P,V,T) ( P , V , T ) Q Q Q heat : energy transferred because of a temperature difference . A path function (depends on how you got there).W W W work : energy transferred by the gas pushing on its surroundings (volume change). Also a path function .
U U U Q Q Q W W W water level (U U U right now .
But "rain in" (Q Q Q W W W processes — you can reach the
same level by lots of rain & lots of outflow, or little of each.
d U dU d U exact differential ; d Q dQ d Q d W dW d W inexact (hence often
written δ Q \delta Q δ Q δ W \delta W δ W
Common mistake Steel-man: "Chemistry usesd U = d Q + d W dU = dQ + dW d U = d Q + d W Why it feels right: Chemists write Δ U = Q + W \Delta U = Q + W Δ U = Q + W The truth: It's the same physics , different definition of W W W
Physics: W W W by the gas → d U = d Q − d W dU = dQ - dW d U = d Q − d W
Chemistry: W W W on the gas → d U = d Q + W on dU = dQ + W_{\text{on}} d U = d Q + W on W by = − W on W_{\text{by}} = -W_{\text{on}} W by = − W on Δ U \Delta U Δ U Fix: Always ask "Whose work — by or on the gas?" before plugging signs.
Common mistake Steel-man: "Compression means I add energy, soW W W Why it feels right: Compressing does pump energy into the gas — feels positive.
The truth: In physics convention W W W by the gas. During compression
d V < 0 dV<0 d V < 0 W = ∫ P d V < 0 W=\int P\,dV < 0 W = ∫ P d V < 0 energy in shows up as − W > 0 -W>0 − W > 0 d U = d Q − d W dU=dQ-dW d U = d Q − d W Fix: Track the sign of d V dV d V
Worked example Example 1 — Heat added, gas expandsA gas absorbs 500 500\, 500 200 200\, 200 Δ U \Delta U Δ U
Identify signs: Q = + 500 Q=+500 Q = + 500 W = + 200 W=+200 W = + 200 Why? Heat goes in → positive; gas pushes piston → positive work.Apply: Δ U = Q − W = 500 − 200 = + 300 \Delta U = Q - W = 500 - 200 = +300\, Δ U = Q − W = 500 − 200 = + 300 Why? Of the 500 J in, 200 J left as work; 300 J stayed → hotter gas.
Worked example Example 2 — Compression with heat releaseA gas is compressed, surroundings do 150 150\, 150 90 90\, 90 Δ U \Delta U Δ U
Signs: work on gas = 150 =150 = 150 W by = − 150 W_{\text{by}} = -150 W by = − 150 Q = − 90 Q=-90 Q = − 90 Why? W W W Apply: Δ U = Q − W = ( − 90 ) − ( − 150 ) = + 60 \Delta U = Q - W = (-90) - (-150) = +60\, Δ U = Q − W = ( − 90 ) − ( − 150 ) = + 60 Why? 150 J pumped in, 90 J leaked out as heat → net +60 J stored.
Worked example Example 3 — Isothermal process (special trick)An ideal gas expands isothermally (T T T Q = 400 Q=400\, Q = 400 W W W
Key fact: For an ideal gas U = U ( T ) U=U(T) U = U ( T ) T T T Δ U = 0 \Delta U=0 Δ U = 0 Why? Internal energy of an ideal gas depends solely on temperature.Apply: 0 = Q − W ⇒ W = Q = 400 0 = Q - W \Rightarrow W = Q = 400\, 0 = Q − W ⇒ W = Q = 400 Why? With no change in U U U all heat absorbed becomes work output.
Worked example Example 4 — Adiabatic processA gas is compressed adiabatically; 200 200\, 200 Δ U \Delta U Δ U
Key fact: Adiabatic ⇒ Q = 0 Q=0 Q = 0 W by = − 200 W_{\text{by}}=-200 W by = − 200 Why? No heat exchange in adiabatic; compression is work on gas.Apply: Δ U = 0 − ( − 200 ) = + 200 \Delta U = 0 - (-200) = +200\, Δ U = 0 − ( − 200 ) = + 200 heats up .
Why? Energy can only come from work; it all stays as internal energy.
Recall Predict before you compute
A gas does 0 0 0 80 80 80 Δ U \Delta U Δ U (then check)
Free expansion into vacuum (W = 0 W=0 W = 0 Q = 0 Q=0 Q = 0 Δ U \Delta U Δ U Δ T \Delta T Δ T
Verify: (1) Δ U = 80 − 0 = + 80 \Delta U = 80 - 0 = +80 Δ U = 80 − 0 = + 80 Δ U = 0 \Delta U=0 Δ U = 0 T T T
Intuition If you remember only this
d U = d Q − d W dU = dQ - dW d U = d Q − d W W W W by the gas = ∫ P d V = \int P\,dV = ∫ P d V U U U T T T Q , W Q,W Q , W Specials: isothermal → Δ U = 0 \Delta U=0 Δ U = 0 Q = 0 Q=0 Q = 0 W = 0 W=0 W = 0 W = P Δ V W=P\Delta V W = P Δ V
"QUWe by" → Δ U = Q − W \Delta U = Q - W Δ U = Q − W W W W by the gas.
Picture: Q ueue (heat lining up to enter) minus the gas W alking out
(pushing the piston out = leaving). What stays inside = Δ U \Delta U Δ U
Recall Feynman: explain to a 12-year-oldImagine your stomach is a balloon of gas. Eating food = heat Q Q Q going in.
That energy either makes you warmer inside (Δ U \Delta U Δ U or lets you
push and lift things (W W W . If you eat 500 calories and burn 200 doing
push-ups, 300 stay stored. That's the First Law: *what you eat = what you store
what you spend.* And "warmth inside" only depends on how hot you are right now,
not on how you got hot — but "food eaten" and "work done" depend on the journey.
State the First Law in physics convention Δ U = Q − W \Delta U = Q - W Δ U = Q − W , where
W W W is work done by the gas.
Why is U U U It depends only on the current state
( P , V , T ) (P,V,T) ( P , V , T ) , not the path taken.
Why are Q Q Q W W W They describe processes; their values depend on how the system changes, not just endpoints.
Derive d W = P d V dW = P\,dV d W = P d V Gas force on piston
= P A =PA = P A ; work
= P A d x = P d V =PA\,dx=P\,dV = P A d x = P d V since
A d x = d V A\,dx=dV A d x = d V .
Sign of W W W Positive (
d V > 0 dV>0 d V > 0 , gas does work on surroundings).
Sign of W W W Negative (
d V < 0 dV<0 d V < 0 ); equivalently surroundings do positive work on gas.
For an isothermal ideal-gas process, Δ U = ? \Delta U=? Δ U = ? Zero, because
U U U depends only on
T T T and
T T T is constant.
For an adiabatic process, which term is zero Q = 0 Q=0 Q = 0 , so
Δ U = − W \Delta U = -W Δ U = − W .
Chemistry writes Δ U = Q + W \Delta U=Q+W Δ U = Q + W No; there
W W W is work done ON the gas
= − W by =-W_\text{by} = − W by , same physics.
Heat 500 500 500 200 200 200 Δ U \Delta U Δ U 500 − 200 = + 300 500-200=+300 500 − 200 = + 300 J.
Intuition Hinglish mein samjhoDekho yaar, First Law of Thermodynamics ka matlab sirf itna hai: energy kahin
se aati nahi, kahin jaati nahi — bas form badalti hai. Agar tum gas ko heat Q Q Q andar se garam ho
jayegi (internal energy U U U piston ko bahar dhakkega yaani
work W W W Δ U = Q − W \Delta U = Q - W Δ U = Q − W Q Q Q W W W Δ U \Delta U Δ U
Sabse zaroori cheez hai sign convention . Physics mein W W W gas
ne jo kaam kiya (work BY gas). Jab gas phailti hai (d V > 0 dV>0 d V > 0 W W W d V < 0 dV<0 d V < 0 W W W Q Q Q Q Q Q Δ U = Q + W \Delta U = Q + W Δ U = Q + W W W W gas par kiya gaya kaam hota hai, physics same hi rehti
hai, bas naam ka farq hai.
Yaad rakhne ke liye special cases ratt lo: isothermal (T T T Δ U = 0 \Delta U = 0 Δ U = 0 U U U Adiabatic mein Q = 0 Q=0 Q = 0 Δ U \Delta U Δ U Isochoric (V V V W = 0 W=0 W = 0 U U U
