Skip to content
Login Register

#volume change

37 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

In an isobaric process, 1.1 moles of an ideal gas expand from 7 L to 14 L at 390 K . What is the work done by the gas? (

**Energy transfer** first law ΔU = Q - W, W includes P-V work, shaft work, electrical work, Q includes conduction Fourier law, convection, radiation Stefan-Boltzmann, distinction important because work is controllable, heat spontaneous from hot to cold, entropy associated with heat not work, explaining why heat engine efficiency

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

View Details
Work Heat Distinction and Energy Transfer Modes

A gas undergoes an adiabatic expansion from 28 L to 84 L , reducing its pressure from 15 atm to 3 atm . What is the valu

**Energy transfer** first law ΔU = Q - W, W includes P-V work, shaft work, electrical work, Q includes conduction Fourier law, convection, radiation Stefan-Boltzmann, distinction important because work is controllable, heat spontaneous from hot to cold, entropy associated with heat not work, explaining why heat engine efficiency

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

View Details
Isochoric Processes and Pressure-Temperature

A gas undergoes an adiabatic expansion from 25 L to 100 L , reducing its pressure from 16 atm to 1 atm . What is the val

**Pressure-temperature relation** at constant volume Gay-Lussac law P ∝ T, for V constant, P₁/T₁ = P₂/T₂, if T doubles from 300 K to 600 K P doubles, e.g., P₁=1 atm at 300 K P₂=2 atm at 600 K, no work done, ΔU = n C_v ΔT = Q. For adiabatic: P₁ V₁^γ = P₂ V₂^γ . 16 × 25^γ = 1 × 100^γ . 16 = ((100)/(25))^γ ⇒ 16 = 4^γ . 4^γ = 2⁴ ⇒ 2²γ = 2⁴ ⇒ 2γ = 4 ⇒ γ = 2 . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

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

View Details
Isobaric and Isothermal Processes Work Calculation

A gas undergoes an adiabatic expansion from 22 L to 66 L , reducing its pressure from 12 atm to 2 atm . What is the valu

**Work calculation** isobaric W = P(V₂ - V₁), e.g., P=1 atm=1.013×10⁵ Pa ΔV=0.01 m³ W=1013 J. Isothermal W = n R T ln(V₂/V₁), for n=1 mol T=300 K V₂/V₁=2 W=1×8.314×300×ln2=1729 J, work done by gas during expansion, area under P-V curve. For adiabatic: P₁ V₁^γ = P₂ V₂^γ . 12 × 22^γ = 2 × 66^γ . 12 / 2 = ((66)/(22))^γ ⇒ 6 = 3^γ . 3^γ = 3¹.63 , γ ≈ 1.63 ≈ 1.67 (standard value from context). Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁),

Ref: NCERT > Physics Book > Thermodynamics > Isobaric and Isothermal Processes Work Calculation

View Details
Ideal Gas Laws and Thermal Properties

A steel cylinder has a volume of 1.5L at 55∘C. What temperature must it be heated to for the volume to increase by 0.005

Given: V0 = 1.5L = 1500cm3, ΔV = 0.0054L = 5.4cm3, αl = 1.2×10−5K−1, T1 = 55∘C. αv = 3αl = 3×1.2×10−5 = 3.6×10−5K−1. ΔV = V0αvΔT⇒5.4 = 1500×3.6×10−5×ΔT. ΔT = 5.41500×3.6×10−5 = 5.40.054 = 100K. T2 = 55+100 = 155∘C.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

View Details
Specific Heat and Calorimetry

A gas at 300K and 1atm occupies 3L. If it is heated to 450K while the volume is adjusted to 4.5L, what is the final pres

Given: T1 = 300K, P1 = 1atm, V1 = 3L, T2 = 450K, V2 = 4.5L. P1V1T1 = P2V2T2. P2 = P1×V1V2×T2T1 = 1×34.5×450300 = 1×23×1.5 = 1atm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1 atm. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

View Details
Fluid Pressure, Pascal's Law and Manometers

What is the primary factor affecting the volume expansion of a gas compared to solids and liquids?

Gases expand much more than solids and liquids due to their higher coefficient of volume expansion, which is temperature-dependent and significantly larger (Section 10.5). As per NCERT, applying relevant law/formula with correct units and sign convention leads to Coefficient of volume expansion. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

View Details