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#temperature drop

3 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

A monatomic ideal gas undergoes an adiabatic expansion, reducing its temperature from 400 K to 300 K . How much work is

**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). Work done: W = (μ R (T₁ - T₂))/(γ - 1) . μ = 1 , R = 8.3 , T₁ = 400 , T₂ = 300 , γ = 1.67 . W = (1 × 8.3 × (400 - 300))/(1.67 - 1) = (8.3 × 100)/(0.67) ≈ 1238.8 J . Using

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

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Isochoric Processes and Pressure-Temperature

A monatomic gas undergoes an adiabatic expansion from 820 K to 410 K with 0.9 moles . What is the work done? ( R = 8.3 J

**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. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 0.9 , R = 8.3 , T₁ = 820 , T₂ = 410 , γ = 1.67 . W = (0.9 × 8.3 × (820 - 410))/(1.67 - 1) = (7.47 × 410)/(0.67) ≈ 4570.15 J ≈ 4570 J . Using first law ΔU = Q - W, W = ∫ P

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

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Zeroth Law Thermal Equilibrium and Quasi-static

What happens to the temperature of an ideal gas during an adiabatic expansion?

**Quasi-static process** infinitely slow, system always near equilibrium, reversible, can be represented as continuous path on P-V diagram, non-quasi-static rapid process non-equilibrium, work W = ∫ P_ext dV, for quasi-static P_ext = P_system, work = ∫ P dV, zeroth law ensures temperature defined throughout quasi-static. In an adiabatic expansion ( Δ Q = 0 ), the gas does work on the surroundings ( W > 0 ), reducing its internal energy ( Δ U = -W ). For an ideal gas, U depends only on temperature, so temperature decreases. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W =

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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