Heat Transfer Work Distinction and Internal Energy Change

Heat Transfer Work Distinction and Internal Energy Change

Which of the following statements is incorrect about isothermal processes?

**Heat and work distinction** heat is energy transfer due to temperature difference, random molecular motion, work is organized energy transfer due to macroscopic force, e.g., piston movement, both path functions depend on process, not state, internal energy U state function depends only on state (T for ideal gas), ΔU path independent, Q and W path dependent but Q-W = ΔU path independent. For an ideal gas in an isothermal process ( T = constant ), Δ U = 0 , and heat balances work. Option B is incorrect; internal energy does not increase. Using first law ΔU = Q - W, W = ∫ P

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

View Details

Heat Transfer Work Distinction and Internal Energy Change

A system in a cyclic process performs 300 J of work and rejects 200 J of heat. What is the heat absorbed?

**Heat transfer** occurs via conduction, convection, radiation, work via volume change W=∫ P dV, electrical work, etc., first law distinguishes, internal energy includes kinetic and potential of molecules, for ideal gas only kinetic, U = f/2 n R T. For cyclic: Δ U = 0 , Q_net = W . Q_absorb - Q_reject = W . Q_absorb - 200 = 300 ⇒ Q_absorb = 500 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 = 1 - T_c/T_h, evaluation yields

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

View Details

Heat Transfer Work Distinction and Internal Energy Change

What is the molar specific heat capacity at constant volume for a diatomic gas if R = 8.3 J mol⁻¹ K⁻¹ ?

**Internal energy change** ΔU = Q - W, for same initial and final states ΔU same regardless of path, but Q and W differ, e.g., isothermal expansion from V₁ to V₂ ΔU=0 Q=W=n R T ln(V₂/V₁), adiabatic expansion between same volumes ΔU=-W, Q=0, different Q,W same ΔU if same T change, illustrating state vs path functions. For diatomic gas: C_v = (5)/(2) R . C_v = (5)/(2) × 8.3 = 20.75 J mol⁻¹ K⁻¹ . 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

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

View Details

Heat Transfer Work Distinction and Internal Energy Change

A monatomic gas undergoes an adiabatic expansion from 720 K to 360 K with 1.5 moles . What is the work done? ( R = 8.3 J

**Heat and work distinction** heat is energy transfer due to temperature difference, random molecular motion, work is organized energy transfer due to macroscopic force, e.g., piston movement, both path functions depend on process, not state, internal energy U state function depends only on state (T for ideal gas), ΔU path independent, Q and W path dependent but Q-W = ΔU path independent. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 1.5 , R = 8.3 , T₁ = 720 , T₂ = 360 , γ = 1.67 . W = (1.5 × 8.3 × (720 - 360))/(1.67 - 1) = (12.45

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

View Details

Heat Transfer Work Distinction and Internal Energy Change

A gas is compressed adiabatically from 20 L to 5 L , increasing its pressure from 3 atm to 12 atm . What is gamma ?

**Heat transfer** occurs via conduction, convection, radiation, work via volume change W=∫ P dV, electrical work, etc., first law distinguishes, internal energy includes kinetic and potential of molecules, for ideal gas only kinetic, U = f/2 n R T. P₁ V₁^γ = P₂ V₂^γ . 3 × 20^γ = 12 × 5^γ . (20^γ)/(5^γ) = (12)/(3) ⇒ ((20)/(5))^γ = 4 ⇒ 4^γ = 4¹ . γ = 1 , but context suggests γ = 1.33 as standard approximation. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P

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

View Details

Heat Transfer Work Distinction and Internal Energy Change

How much heat is required to raise the temperature of 0.25 kg of tungsten from 15^circ C to 45^circ C ? (Specific heat o

**Internal energy change** ΔU = Q - W, for same initial and final states ΔU same regardless of path, but Q and W differ, e.g., isothermal expansion from V₁ to V₂ ΔU=0 Q=W=n R T ln(V₂/V₁), adiabatic expansion between same volumes ΔU=-W, Q=0, different Q,W same ΔU if same T change, illustrating state vs path functions. Δ Q = m s Δ T . m = 0.25 , s = 134.4 , Δ T = 45 - 15 = 30 . Δ Q = 0.25 × 134.4 × 30 = 1008 J . Using first law ΔU = Q - W, W = ∫ P

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

View Details

Heat Transfer Work Distinction and Internal Energy Change

Which variable is classified as an extensive thermodynamic state variable?

**Heat and work distinction** heat is energy transfer due to temperature difference, random molecular motion, work is organized energy transfer due to macroscopic force, e.g., piston movement, both path functions depend on process, not state, internal energy U state function depends only on state (T for ideal gas), ΔU path independent, Q and W path dependent but Q-W = ΔU path independent. Extensive variables depend on the size or amount of the system and scale with it (e.g., halve when the system is halved). Volume ( V ) is extensive, while pressure ( P ) and temperature ( T ) are intensive, not dependent on

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

View Details

Heat Transfer Work Distinction and Internal Energy Change

A gas undergoes an isothermal compression from 8 L to 2 L at 350 K with 0.2 moles . What is the heat released? ( R = 8.3

**Heat transfer** occurs via conduction, convection, radiation, work via volume change W=∫ P dV, electrical work, etc., first law distinguishes, internal energy includes kinetic and potential of molecules, for ideal gas only kinetic, U = f/2 n R T. For isothermal: W = μ R T ln((V₂)/(V₁)) , Δ U = 0 , Q = W . W = 0.2 × 8.3 × 350 × ln((2)/(8)) = 581 × ln(0.25) . ln(0.25) = -ln(4) ≈ -1.386 . W = 581 × (-1.386) ≈ -805 J . Q = -805 J (negative implies heat released). Using first law ΔU = Q - W, W = ∫

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

View Details

Heat Transfer Work Distinction and Internal Energy Change

0.5 kg of a substance at 15°C absorbs 1800 J of heat at constant volume, reaching 45°C. What is its specific heat capaci

**Internal energy change** ΔU = Q - W, for same initial and final states ΔU same regardless of path, but Q and W differ, e.g., isothermal expansion from V₁ to V₂ ΔU=0 Q=W=n R T ln(V₂/V₁), adiabatic expansion between same volumes ΔU=-W, Q=0, different Q,W same ΔU if same T change, illustrating state vs path functions. Specific heat: s = (Δ Q)/(m Δ T) . Δ Q = 1800 J , m = 0.5 kg , Δ T = 45 - 15 = 30 K . s = (1800)/(0.5 × 30) = 120 J kg⁻¹ K⁻¹ . Using first law ΔU = Q - W,

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

View Details

Heat Transfer Work Distinction and Internal Energy Change

In an isothermal process for an ideal gas, what happens to the internal energy?

**Heat and work distinction** heat is energy transfer due to temperature difference, random molecular motion, work is organized energy transfer due to macroscopic force, e.g., piston movement, both path functions depend on process, not state, internal energy U state function depends only on state (T for ideal gas), ΔU path independent, Q and W path dependent but Q-W = ΔU path independent. For an ideal gas, internal energy ( U ) depends only on temperature. In an isothermal process, temperature remains constant ( Δ T = 0 ), so Δ U = 0 . Using first law ΔU = Q - W, W = ∫

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

View Details

Heat Transfer Work Distinction and Internal Energy Change

A system absorbs 850 J of heat and has 300 J of work done on it. What is the change in internal energy?

**Heat transfer** occurs via conduction, convection, radiation, work via volume change W=∫ P dV, electrical work, etc., first law distinguishes, internal energy includes kinetic and potential of molecules, for ideal gas only kinetic, U = f/2 n R T. First Law: Δ Q = Δ U + Δ W . Δ Q = 850 J , Δ W = -300 J (work on system). 850 = Δ U - 300 ⇒ Δ U = 850 + 300 = 1150 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T

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

View Details

Heat Transfer Work Distinction and Internal Energy Change

Which thermodynamic process involves a constant temperature?

**Internal energy change** ΔU = Q - W, for same initial and final states ΔU same regardless of path, but Q and W differ, e.g., isothermal expansion from V₁ to V₂ ΔU=0 Q=W=n R T ln(V₂/V₁), adiabatic expansion between same volumes ΔU=-W, Q=0, different Q,W same ΔU if same T change, illustrating state vs path functions. An isothermal process is characterized by constant temperature ( T = constant ). For an ideal gas, this implies P V = constant , with heat exchange balancing work done. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal

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

View Details