First Law of Thermodynamics Applications

First Law of Thermodynamics Applications

How much heat is required to vaporize 0.5 g of water at 100^circ C and 1 atm ? (Latent heat = 2256 J/g )

**First law of thermodynamics** ΔU = Q - W, ΔU internal energy change (J), Q heat added to system (J), W work done by system (J), sign convention physics Q positive when added, W positive when done by system, energy conservation, for isochoric W=0 ΔU=Q, for adiabatic Q=0 ΔU=-W, for isothermal ΔU=0 Q=W, for cyclic ΔU=0 Q_net=W_net. Δ Q = m L . m = 0.5 , L = 2256 . Δ Q = 0.5 × 2256 = 1128 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T

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First Law of Thermodynamics Applications

How many joules are equivalent to 250 cal of heat? (1 cal = 4.186 J )

**Internal energy** state function depends only on temperature for ideal gas, U = f/2 n R T, change ΔU = n C_v ΔT, first law connects heat, work, internal energy, for expansion work done by gas positive, compression work done on gas negative, heat added positive. Heat in J = Heat in cal × 4.186 . 250 × 4.186 = 1046.5 J ≈ 1047 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 1047

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First Law of Thermodynamics Applications

0.1 kg of a substance absorbs 1200 J of heat, increasing its temperature from 20°C to 50°C. What is its specific heat ca

**First law applications** for isobaric W = P ΔV, Q = n C_p ΔT, ΔU = n C_v ΔT, for isothermal ideal gas ΔU=0 Q=W=n R T ln(V₂/V₁), for adiabatic Q=0 W= -ΔU = (P₁V₁ - P₂V₂)/(γ-1), for isochoric W=0 ΔU=Q=n C_v ΔT, enabling calculation of Q,W,ΔU for any process. Specific heat: s = (Δ Q)/(m Δ T) . Given Δ Q = 1200 J , m = 0.1 kg , Δ T = 50 - 20 = 30 K . s = (1200)/(0.1 × 30) = 400 J kg⁻¹ K⁻¹ . Using first law ΔU = Q - W, W = ∫ P dV,

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First Law of Thermodynamics Applications

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

**First law of thermodynamics** ΔU = Q - W, ΔU internal energy change (J), Q heat added to system (J), W work done by system (J), sign convention physics Q positive when added, W positive when done by system, energy conservation, for isochoric W=0 ΔU=Q, for adiabatic Q=0 ΔU=-W, for isothermal ΔU=0 Q=W, for cyclic ΔU=0 Q_net=W_net. For diatomic gas: C_v = (5)/(2) R , C_p = C_v + R = (7)/(2) R . C_p = (7)/(2) × 8.3 = 29.05 J mol⁻¹ K⁻¹ ≈ 29.1 J mol⁻¹ K⁻¹ . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W

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First Law of Thermodynamics Applications

Which of the following statements is correct about an isothermal process for an ideal gas?

**Internal energy** state function depends only on temperature for ideal gas, U = f/2 n R T, change ΔU = n C_v ΔT, first law connects heat, work, internal energy, for expansion work done by gas positive, compression work done on gas negative, heat added positive. In an isothermal process ( T = constant ), the internal energy of an ideal gas ( U , temperature-dependent) remains constant ( Δ U = 0 ), and heat supplied equals work done ( Δ Q = Δ W ), making option C correct. Using first law ΔU = Q - W, W = ∫ P dV, isobaric

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First Law of Thermodynamics Applications

Which of the following statements is incorrect about the Zeroth Law of Thermodynamics?

**First law applications** for isobaric W = P ΔV, Q = n C_p ΔT, ΔU = n C_v ΔT, for isothermal ideal gas ΔU=0 Q=W=n R T ln(V₂/V₁), for adiabatic Q=0 W= -ΔU = (P₁V₁ - P₂V₂)/(γ-1), for isochoric W=0 ΔU=Q=n C_v ΔT, enabling calculation of Q,W,ΔU for any process. The Zeroth Law defines temperature through thermal equilibrium but does not address heat flow direction (Second Law) or energy conservation (First Law). Option C is incorrect. 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 =

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First Law of Thermodynamics Applications

How much heat is required to vaporize 0.7 g of water at 100^circ C and 1 atm ? (Latent heat = 2256 J/g )

**First law of thermodynamics** ΔU = Q - W, ΔU internal energy change (J), Q heat added to system (J), W work done by system (J), sign convention physics Q positive when added, W positive when done by system, energy conservation, for isochoric W=0 ΔU=Q, for adiabatic Q=0 ΔU=-W, for isothermal ΔU=0 Q=W, for cyclic ΔU=0 Q_net=W_net. Δ Q = m L . m = 0.7 , L = 2256 . Δ Q = 0.7 × 2256 = 1579.2 J ≈ 1579 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W =

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First Law of Thermodynamics Applications

How many calories are equivalent to 836 J of heat? (1 cal = 4.186 J )

**Internal energy** state function depends only on temperature for ideal gas, U = f/2 n R T, change ΔU = n C_v ΔT, first law connects heat, work, internal energy, for expansion work done by gas positive, compression work done on gas negative, heat added positive. Heat in cal = Heat in J4.186 . (836)/(4.186) ≈ 199.71 ≈ 200 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 200 cal, consistent with thermodynamic laws and energy conservation.

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First Law of Thermodynamics Applications

An ideal gas expands isothermally at 380 K from 5 L to 15 L with 0.25 moles . What is the work done by the gas? ( R = 8.

**First law applications** for isobaric W = P ΔV, Q = n C_p ΔT, ΔU = n C_v ΔT, for isothermal ideal gas ΔU=0 Q=W=n R T ln(V₂/V₁), for adiabatic Q=0 W= -ΔU = (P₁V₁ - P₂V₂)/(γ-1), for isochoric W=0 ΔU=Q=n C_v ΔT, enabling calculation of Q,W,ΔU for any process. For isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.25 , R = 8.3 , T = 380 , V₂ = 15 , V₁ = 5 . W = 0.25 × 8.3 × 380 × ln((15)/(5)) = 788.5 × ln(3) . ln(3) ≈ 1.0986 , W ≈ 788.5 × 1.0986 ≈ 866 J

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First Law of Thermodynamics Applications

A system absorbs 690 J of heat and has 210 J of work done on it. What is the change in internal energy?

**First law of thermodynamics** ΔU = Q - W, ΔU internal energy change (J), Q heat added to system (J), W work done by system (J), sign convention physics Q positive when added, W positive when done by system, energy conservation, for isochoric W=0 ΔU=Q, for adiabatic Q=0 ΔU=-W, for isothermal ΔU=0 Q=W, for cyclic ΔU=0 Q_net=W_net. First Law: Δ Q = Δ U + Δ W . Δ Q = 690 , Δ W = -210 (work on system). 690 = Δ U - 210 ⇒ Δ U = 690 + 210 = 900 J . Using first law ΔU = Q - W,

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First Law of Thermodynamics Applications

How much heat is required to raise the temperature of 0.25 kg of tungsten from 25^circ C to 55^circ C ? (Specific heat o

**Internal energy** state function depends only on temperature for ideal gas, U = f/2 n R T, change ΔU = n C_v ΔT, first law connects heat, work, internal energy, for expansion work done by gas positive, compression work done on gas negative, heat added positive. Δ Q = m s Δ T . m = 0.25 , s = 134.4 , Δ T = 55 - 25 = 30 . Δ Q = 0.25 × 134.4 × 30 = 1008 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n

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First Law of Thermodynamics Applications

0.2 moles of an ideal gas expand isothermally at 350 K from 4 L to 10 L. What is the heat absorbed? ( R = 8.3 J mol⁻¹ K⁻

**First law applications** for isobaric W = P ΔV, Q = n C_p ΔT, ΔU = n C_v ΔT, for isothermal ideal gas ΔU=0 Q=W=n R T ln(V₂/V₁), for adiabatic Q=0 W= -ΔU = (P₁V₁ - P₂V₂)/(γ-1), for isochoric W=0 ΔU=Q=n C_v ΔT, enabling calculation of Q,W,ΔU for any process. Isothermal: Δ U = 0 , Δ Q = Δ W = μ R T ln((V₂)/(V₁)) . μ = 0.2 , T = 350 , V₂ = 10 , V₁ = 4 . Δ Q = 0.2 × 8.3 × 350 × ln((10)/(4)) = 581 × 0.916 ≈ 532 J . Using first law ΔU

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