First law of thermodynamics — dU = dQ − dW, sign conventions: Question bank

1.7.13 · D5 · Physics › Thermodynamics › First law of thermodynamics — dU = dQ − dW, sign conventions

Definition Dono sign conventions, side by side (dono signs yaad karo!)

Quantity	Positive $(+)$ jab…	Negative $(-)$ jab…
Heat $Q$	heat gas ke andar jaati hai (gas absorb karti hai)	heat gas ke bahar jaati hai (gas release karti hai)
Work $W$ (work by gas)	gas expand karti hai, $d V > 0$ (energy gas se bahar jaati hai)	gas compress hoti hai, $d V < 0$ (energy gas mein aati hai)

Toh $Δ U = Q - W$ : heat in " $+$ " hai, work out " $-$ " hai.

Figures — path dependence aur sign zones kaisi dikhti hain

Work, $P$ – $V$ curve ke neeche ka area hai; same endpoints ke beech do paths alag-alag areas enclose karti hain — isliye $Q$ aur $W$ path functions hain.

Figure — First law of thermodynamics — dU = dQ − dW, sign conventions

Sign-convention zones: expansion (area sweeping right, $W > 0$ ) versus compression (swept left, $W < 0$ ); heat in $(Q > 0)$ versus heat out $(Q < 0$ ).

$Q$ path-dependent kyun hai: do routes A→B ek area enclose karti hain = net work, aur $Q = Δ U + W$ se woh alag $W$ ek alag $Q$ force karta hai chahe $Δ U$ identical ho.

True or false — justify karo

Heat energy ka ek form hai jo gas ke andar stored hoti hai.

False. Heat energy hai in transit, jo temperature difference se driven hoti hai; ek baar andar aane par yeh internal energy

U

ban jaati hai — tum gas ke andar "the heat" point nahi kar sakte.

Agar do processes same

(P, V, T)

par start aur end karti hain, toh unhe same

Q

transfer karna chahiye.

False.

Q

ek path function hai. Sirf

Δ U

(state function) endpoints se fix hota hai;

Q

aur

W

bahut alag ho sakte hain jabki

Q - W

equal rehta hai — upar wala two-path figure dekho.

Agar do processes ke endpoints same hain, toh unka

Δ U

bhi same hoga.

True.

U

ek state function hai, isliye

Δ U = U_{final} - U_{initial}

sirf endpoints par depend karta hai, kabhi path par nahi.

Physics convention mein, gas ko compress karna positive

W

deta hai.

False.

W

gas ke dwara kiya gaya work hai; compression ka matlab

d V < 0

hai, toh

W < 0

. Jo energy tum pump karte ho woh

Δ U = Q - W

mein

- W > 0

ke roop mein dikhti hai.

Ideal gas mein, heat add karne se temperature hamesha badhti hai.

False. Isothermal expansion mein gas heat absorb karti hai phir bhi

T

constant rehta hai — saari heat phir se work ke roop mein nikal jaati hai (

Δ U = 0

First Law batata hai ki ek process kis direction mein jaayegi.

False. First Law sirf energy bookkeeping enforce karta hai; direction (irreversibility) Second law of thermodynamics ka kaam hai.

Δ U = Q - W

sirf ideal gases ke liye hold karta hai.

False. Yeh pure energy conservation hai aur kisi bhi closed system ke liye hold karta hai — solids, liquids, real gases bhi. Sirf shortcut

Δ U = n C_{V} Δ T

ke liye ideal-gas assumption chahiye.

Vacuum mein free expansion mein, gas positive work karti hai kyunki woh expand karti hai.

False. Work

= \int P_{ext} d V

; external pressure zero hai (vacuum), toh

W = 0

chahe volume badh raha ho.

First Law mein sirf ek hi tarah ka work hota hai — pressure–volume ("

P d V

") work.

False.

W

saara work hai jo system karta hai: battery par electrical work, magnetic work, stretched film par surface-tension work, shaft work, etc. Simple gas ke liye sirf channel

P d V

hai, lekin law

Δ U = Q - W_{total}

har mode cover karta hai.

Chemistry ka

Δ U = Q + W

physics ke

Δ U = Q - W

se contradict karta hai.

False. Same physics hai, alag

W

: chemistry ka

W

gas par kiya gaya work hai

= - W_{by}

, toh dono identical

Δ U

dete hain.

Ek adiabatic process fast honi chahiye.

Zyaadatar sahi hai, lekin asli requirement $Q = 0$ hai. Fast processes heat flow ke liye time nahi deti, aur perfect insulation bhi

Q = 0

deta hai — dono tarike se adiabatic milta hai.

Work tab hota hai jab piston jo actually push kar raha hai use — surroundings — ke against push ho. Sirf slow (reversible) process mein gas pressure har instant par external pressure ke equal hota hai; sudden expansion mein gas pressure

P_{ext}

se zyada ho sakta hai, aur real work smaller

\int P_{ext} d V

hota hai.

First Law yeh predict karne ke liye useless kyun hai ki ice spontaneously pighlayi?

Yeh sirf check karta hai ki energy balance ho; yeh energy-conserving processes ko "backwards" chalane se nahi rokta. Spontaneity Second law of thermodynamics mein entropy se governed hoti hai.

Edge cases

Vacuum mein ideal gas ki free expansion:

Q

W

Δ U

Δ T

kya hain?

W = 0

(external pressure zero hai),

Q = 0

(fast, insulated), toh

Δ U = 0

aur

Δ T = 0

— temperature unchanged rehti hai chahe volume jump kare.

Ek cyclic process apni start par wapas aati hai: ek poore cycle mein

Δ U

kya hai?

Zero, kyunki

U

ek state function hai aur start aur end states identical hain; isliye cycle mein

Q_{net} = W_{net}

—

P

–

V

diagram par loop se enclosed area.

Isochoric (constant-volume) heating: kaunsa term vanish hota hai aur kya bachta hai?

W = \int P d V = 0

kyunki

d V = 0

, toh

Δ U = Q

— heat ka har joule internal energy ban jaata hai.

Adiabatic process mein

Q = 0

: kya iska matlab

Δ U = 0

hai?

Nahi.

Q = 0

Δ U = - W

milta hai, toh internal energy exactly minus work done by gas ke barabar change hoti hai — compression par garam hota hai, expansion par thanda.

Ek gas constant temperature par expand karti hai lekin real gas hai jisme attractions hain: kya

Δ U

abhi bhi zero hai?

Zaruri nahi. Real gases mein intermolecular potential energy hoti hai jo volume ke saath change karti hai, toh

U

bhi

V

par depend karta hai; constant

T

par bhi

Δ U

nonzero ho sakta hai.

System ke andar ek battery electrical energy deliver karti hai: kya First Law abhi bhi hold karta hai?

Haan — lekin electrical work ko work mode ke roop mein count karna hoga:

Δ U = Q - (W_{P V} + W_{elec} + \dots)

. Law ne kabhi nahi assume kiya ki work sirf

P d V

hai.

Zero heat aur zero work (

Q = 0

W = 0

): tum kya conclude kar sakte ho?

Δ U = 0

— internal energy (aur ideal gas ke liye, temperature) unchanged hai; state ki energy completely isolated hai.

Ideal gas ki constant-pressure heating:

Q

Δ U

se zyada kyun hota hai?

Kyunki gas expand bhi karti hai aur work

W = P Δ V > 0

karta hai, toh

Q = Δ U + W

; extra heat work ke liye pay karti hai (isliye

C_{P} > C_{V}

hota hai, dekho Heat capacities Cp and Cv).

Gas compress karte ho aur woh utni hi heat bahar shed karti hai jitna work put in kiya:

Δ U

kya hai?

Zero. Gas par work energy badhata hai, heat loss equal amount remove karti hai:

Δ U = Q - W = (- x) - (- x) = 0

Connections

Work done in thermodynamic processes (PV diagrams)
Isothermal, adiabatic, isobaric, isochoric processes
Internal energy and degrees of freedom
Heat capacities Cp and Cv
Conservation of energy (mechanics)
Second law of thermodynamics

Question bank — First law of thermodynamics — dU = dQ − dW, sign conventions

Figures — path dependence aur sign zones kaisi dikhti hain

True or false — justify karo

Error dhundo

Why questions

Edge cases

Connections