Zeroth Law Thermal Equilibrium and Quasi-static

Zeroth Law Thermal Equilibrium and Quasi-static

A gas undergoes an adiabatic compression with 400 J of work done on it. What is the change in internal energy?

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. Adiabatic: Δ Q = 0 , Δ U = -Δ W . Work on system: Δ W = -400 J . Δ U = -(-400) = 400 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 400 J, consistent

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Zeroth Law Thermal Equilibrium and Quasi-static

Which of the following statements is correct about internal energy?

**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. Internal energy ( U ) is a state function, dependent on the system’s state (e.g., temperature for an ideal gas), not the path taken. Option A is correct. 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

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Zeroth Law Thermal Equilibrium and Quasi-static

A system absorbs 800 J of heat and has 350 J of work done on it. What is the change in internal energy?

**Zeroth law of thermodynamics** if two systems A and B each in thermal equilibrium with third C, then A and B in equilibrium with each other, defines temperature as property that is same for systems in thermal equilibrium, basis for thermometer, temperature scale, thermal equilibrium means no net heat flow, same temperature. First Law: Δ Q = Δ U + Δ W . Δ Q = 800 , Δ W = -350 (work done on system). 800 = Δ U - 350 ⇒ Δ U = 800 + 350 = 1150 J . Using first law ΔU = Q - W, W = ∫ P

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Zeroth Law Thermal Equilibrium and Quasi-static

An ideal gas expands isothermally at 540 K from 10 L to 30 L with 0.3 moles . What is the work done by the gas? ( R = 8.

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. For isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.3 , R = 8.3 , T = 540 , V₂ = 30 , V₁ = 10 . W = 0.3 × 8.3 × 540 × ln((30)/(10)) = 1344.6 × ln(3) . ln(3) ≈ 1.0986 , W ≈ 1344.6 × 1.0986 ≈ 1477 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric

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Zeroth Law Thermal Equilibrium and Quasi-static

A diatomic gas undergoes an adiabatic expansion from 860 K to 430 K with 0.5 moles . What is the work done? ( R = 8.3 J

**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. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 0.5 , R = 8.3 , T₁ = 860 , T₂ = 430 , γ = 1.4 . W = (0.5 × 8.3 × (860 - 430))/(1.4 - 1) = (4.15 × 430)/(0.4) = 4467.5 J ≈ 4468 J . Using first law ΔU = Q - W,

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Zeroth Law Thermal Equilibrium and Quasi-static

A gas at 4 atm and 300 K in a 5 L container is compressed isothermally to 2 L. What is the work done on the gas? ( R = 8

**Zeroth law of thermodynamics** if two systems A and B each in thermal equilibrium with third C, then A and B in equilibrium with each other, defines temperature as property that is same for systems in thermal equilibrium, basis for thermometer, temperature scale, thermal equilibrium means no net heat flow, same temperature. Isothermal: W = μ R T ln((V₂)/(V₁)) . P₁ V₁ = μ R T ⇒ 4 × 5 = μ × 8.3 × 300 ⇒ μ = (20)/(2490) ≈ 0.008 mol . W = 0.008 × 8.3 × 300 × ln((2)/(5)) = 19.92 × (-0.916) ≈ -18.25 J (work by gas negative). Work

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Zeroth Law Thermal Equilibrium and Quasi-static

How many calories are equivalent to 2093 J of heat? (1 cal = 4.186 J )

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. Heat in cal = Heat in J4.186 . (2093)/(4.186) ≈ 500 cal . 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 500 cal, consistent with thermodynamic laws and energy conservation.

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Zeroth Law Thermal Equilibrium and Quasi-static

What is the primary implication of the First Law of Thermodynamics?

**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. The First Law of Thermodynamics is a statement of energy conservation: Δ Q = Δ U + Δ W . It implies that the total energy supplied to a system (as heat) equals the increase in internal energy plus the work done by the system. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W

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Zeroth Law Thermal Equilibrium and Quasi-static

A system absorbs 920 J of heat and performs 280 J of work. What is the change in internal energy?

**Zeroth law of thermodynamics** if two systems A and B each in thermal equilibrium with third C, then A and B in equilibrium with each other, defines temperature as property that is same for systems in thermal equilibrium, basis for thermometer, temperature scale, thermal equilibrium means no net heat flow, same temperature. First Law: Δ Q = Δ U + Δ W . Δ Q = 920 , Δ W = 280 (work by system). 920 = Δ U + 280 ⇒ Δ U = 920 - 280 = 640 J . Using first law ΔU = Q - W, W = ∫ P dV,

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Zeroth Law Thermal Equilibrium and Quasi-static

A system in a cyclic process absorbs 980 J of heat and performs 420 J of work. What is the heat rejected?

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. For cyclic: Δ U = 0 , Q_net = W . Q_absorb - Q_reject = W . 980 - Q_reject = 420 ⇒ Q_reject = 980 - 420 = 560 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

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Zeroth Law Thermal Equilibrium and Quasi-static

What is the change in internal energy when 1 mole of an ideal gas is heated from 300 K to 350 K at constant volume? ( C_

**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. Δ U = μ C_v Δ T . μ = 1 , C_v = 20.8 , Δ T = 350 - 300 = 50 . Δ U = 1 × 20.8 × 50 = 1040 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R

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Zeroth Law Thermal Equilibrium and Quasi-static

A system releases 790 J of heat and has 310 J of work done on it. What is the change in internal energy?

**Zeroth law of thermodynamics** if two systems A and B each in thermal equilibrium with third C, then A and B in equilibrium with each other, defines temperature as property that is same for systems in thermal equilibrium, basis for thermometer, temperature scale, thermal equilibrium means no net heat flow, same temperature. First Law: Δ Q = Δ U + Δ W . Δ Q = -790 (heat released), Δ W = -310 (work on system). -790 = Δ U - 310 ⇒ Δ U = -790 + 310 = -480 J . Using first law ΔU = Q - W, W = ∫ P

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