NEB Class 12 • Physics • Worked Practice

First Law of Thermodynamics Practice Set: NEB 12 Physics

Attempt first, draw heat/work arrows, name the process, read P–V area and check whether the final internal-energy change makes physical sense.

Sign-convention and ledger problems
Process and P–V graph calculations
Cycle, path comparison and independent transfer

Define the system and sign convention before balancing heat, work, internal energy and entropy.

Worked set A

Four direct energy ledgers

1. Gas absorbs 800 J and does 300 J work: ΔU=800−300=500 J. 2. Gas loses 120 J while 400 J work is done on it: Q=−120, W=−400, ΔU=280 J. 3. Rigid container receives 650 J: W=0, ΔU=650 J. 4. Adiabatic expansion does 240 J: Q=0, ΔU=−240 J.

For each, write the energy story. In case 2 compression adds more energy than cooling removes. Check signs with OpenStax First Law.

Reverse problem

If ΔU=+350 J and gas does 150 J work, Q=ΔU+W=500 J must enter.

Worked set B

Pressure–volume work and path dependence

A gas expands at constant 1.5×10⁵ Pa by 4.0×10⁻³ m³. W=600 J. If ΔU=250 J, Q=850 J. If the same endpoints are connected by another path with W=900 J, ΔU remains 250 J but Q becomes 1150 J.

A linear P–V path changes pressure from 3.0×10⁵ to 1.0×10⁵ Pa while volume rises by 2.0×10⁻³ m³. Work is trapezoid area, average pressure times ΔV: [(3.0+1.0)/2×10⁵](2.0×10⁻³)=400 J.

The OpenStax work/heat section explains dW=P dV. Unit check: Pa m³=J.

Worked set C

Rectangular cycle and direction

A rectangular clockwise P–V cycle spans pressures 1.0×10⁵ to 3.0×10⁵ Pa and volumes 2.0×10⁻³ to 5.0×10⁻³ m³. Net work is enclosed area: ΔPΔV=(2.0×10⁵)(3.0×10⁻³)=600 J positive. Over a cycle ΔU=0, so Qnet=Wnet=600 J.

If direction reverses, net work becomes −600 J and net heat is −600 J. Individual legs can have heat/work values, but only state return guarantees total ΔU zero. Review cyclic processes.

Process contrast

Isochoric leg has zero boundary work. Constant-pressure expansion/compression legs provide the rectangle’s horizontal-area contributions.

Independent set

Solve ten transfer tasks

Q=500 J, W=200 J: find ΔU.
Q=−80 J, W=−250 J: find ΔU.
Rigid vessel loses 300 J: find ΔU.
Adiabatic compression adds 450 J work: find ΔU.
Constant P=2×10⁵ Pa expansion ΔV=0.004 m³: find W.
Given ΔU and W, solve Q.
Compare two P–V paths between identical states.
Find trapezoid work for linear pressure change.
Find rectangular-cycle net work and heat.
Explain why isothermal ideal-gas expansion needs heat input.

Answers must include arrows, process condition and a sentence check. Use the study guide only after attempting.

Error protocol

Retest signs and paths separately

Classify system, sign, process, path/area, unit, state-function or algebra errors. Redo, then retest with reversed energy flow. For tuition call 9846662070; the MKS panel supports post-Grade-12 preparation and pre-counselling.

Frequently asked questions

Questions about the First Law practice set

What sign convention does this guide use?

Q is positive into the system and W is positive when done by the system, so ΔU=Q−W.

Is heat stored inside a gas?

Heat is energy transfer due to temperature difference; internal energy is a state property.

Why is work path dependent?

Boundary work depends on the pressure–volume path, represented by area under the P–V curve.

What happens in an isochoric process?

Volume does not change, so boundary work is zero and ΔU=Q under this convention.

What happens over a complete cycle?

The system returns to its initial state, so ΔU=0 and net Q equals net W.

Where can I get Thermodynamics tuition?

Call 9846662070 for current KTM Tuition online or physical schedules.

Checked sources

References and related learning

Rebuild concepts with the topic guide. Sources were checked on 2 August 2026; follow current CDC, NEB and school notices if requirements change.

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