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#net work

5 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

A system in a cyclic process absorbs 940 J of heat and rejects 360 J . What is the net work done?

**Work and heat** both energy transfer modes, work organized, e.g., lifting weight, compressing gas, electrical current, heat random due to temperature difference, work can be completely converted to heat via friction, but heat cannot be completely converted to work (second law), energy transfer modes include work (mechanical, electrical) and heat (conduction, convection, radiation). For cyclic: Δ U = 0 , Q_net = W . Q_net = Q_absorb - Q_reject = 940 - 360 = 580 J . W = 580 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

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Isochoric Processes and Pressure-Temperature

A system in a cyclic process absorbs 1020 J of heat and rejects 380 J . What is the net work done?

**Isochoric process** constant volume ΔV=0, work W=0, first law ΔU = Q, all heat goes to internal energy, P/T = constant from ideal gas law P V = n R T at constant V, pressure proportional to temperature, P₁/T₁ = P₂/T₂, e.g., heating gas in rigid container pressure rises proportionally to T. For cyclic: Δ U = 0 , Q_net = W . Q_net = Q_absorb - Q_reject = 1020 - 380 = 640 J . W = 640 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

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Isochoric Processes and Pressure-Temperature

A gas undergoes a cyclic process where 600 J of heat is absorbed. What is the net work done by the gas?

**Isochoric work** zero because dV=0, so W=∫ P dV=0, internal energy change equals heat added, Q = n C_v ΔT, C_v molar specific heat at constant volume, for monatomic 3/2 R, for diatomic 5/2 R, temperature change directly from heat input. For a cyclic process, Δ U = 0 . Δ Q = Δ U + Δ W ⇒ 600 = 0 + Δ W . Δ W = 600 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η =

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

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Isochoric Processes and Pressure-Temperature

A system in a cyclic process absorbs 900 J of heat and rejects 400 J . What is the net work done?

**Isochoric process** constant volume ΔV=0, work W=0, first law ΔU = Q, all heat goes to internal energy, P/T = constant from ideal gas law P V = n R T at constant V, pressure proportional to temperature, P₁/T₁ = P₂/T₂, e.g., heating gas in rigid container pressure rises proportionally to T. For cyclic process: Δ U = 0 , W = Q_net . Q_net = Q_absorb - Q_reject = 900 - 400 = 500 J . W = 500 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

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