Skip to content
Login Register

#Cp

9 public questions tagged with this topic.

Heat Transfer Work Distinction and Internal Energy Change

Which of the following statements is correct about C_p and C_v for an ideal gas?

**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, C_p > C_v because at constant pressure, heat supplies both internal energy increase and work ( C_p = C_v + R ), while at constant volume, heat only increases internal energy. Option A is correct. Using first law ΔU

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

View Details
First Law of Thermodynamics Applications

In an isobaric process, 0.7 moles of gas expand from 320 K to 400 K . What is the heat supplied if C_p = 29.1 J mol⁻¹ K⁻

**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 = μ C_p Δ T . μ = 0.7 , C_p = 29.1 , Δ T = 400 - 320 = 80 . Δ Q = 0.7 × 29.1 × 80 = 1632 J . Using first law ΔU = Q - W, W = ∫

Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications

View Details
Collision Frequency and Mean Free Path Variation

The molar specific heat at constant pressure for a polyatomic gas with 2 vibrational modes is: (R = 8.31 J mol⁻¹ K⁻¹)

**Mean free path variation** λ ∝1/n ∝1/P at constant T, λ ∝ T/P, temperature increase increases λ because n decreases at constant P, but also v increases, overall λ ∝ T/P, for gas at 2 atm λ=4×10⁻⁷ m, at 4 atm λ=2×10⁻⁷ m halves when pressure doubles, as n doubles. Polyatomic gas: 3 translational + 3 rotational + 2 vibrational modes.Total degrees of freedom = 3 + 3 + 2 × 2 = 10.C_v = 5R, C_p = C_v + R = 6R = 6 × 8.31 = 49.86 J mol⁻¹ K⁻¹ . Substituting values gives 49.86 J mol⁻¹ K⁻¹, which matches expected kinetic theory

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Collision Frequency and Mean Free Path Variation

View Details
Molecular Mass Density and Ideal Gas Equation

A gas has a C_p of 20.8 J mol⁻¹ K⁻¹. What is the ratio of specific heats (gamma)? (R = 8.31 J mol⁻¹ K⁻¹)

**Molecular mass and density** relation ρ = P M/(R T) allows density calculation, ideal gas law also P = n k_B T where n number density, molecular mass determines mass per molecule, density increases with pressure and decreases with temperature, inverse T dependence. C_v = C_p - R = 20.8 - 8.31 = 12.49 J mol⁻¹ K⁻¹ ≈ 12.5.γ = (C_p)/(C_v) = (20.8)/(12.5) ≈ 1.66 ≈ 1.67. Substituting values gives 1.67, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Molecular Mass Density and Ideal Gas Equation

View Details
Degrees of Freedom and Molar Specific Heat

What is the ratio of C_p to C_v for a gas with 3 translational and 2 rotational degrees of freedom?

**Specific heat relation** C_p - C_v = R for ideal gas per mole, Mayer's relation, due to work done at constant pressure, degrees of freedom include translational, rotational, vibrational, each quadratic term contributes ½ R to C_v. Degrees of freedom = 3 + 2 = 5.C_v = (5)/(2) R, C_p = C_v + R = (5)/(2) R + R = (7)/(2) R.γ = (C_p)/(C_v) = (7)/(2) R(5)/(2) R = (7)/(5) = 1.4. Substituting values gives 1.4, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Degrees of Freedom and Molar Specific Heat

View Details
Mean Free Path and Molecular Diameter

A gas has a C_p of 35.4 J mol⁻¹ K⁻¹. What is the ratio of specific heats (gamma)? (R = 8.31 J mol⁻¹ K⁻¹)

**Molecular diameter from mean free path** uses λ =1/(√2 n π d²), solving d = √(1/(√2 n π λ)). At higher pressure n ∝ P, λ ∝1/P, so doubling P halves λ, illustrating pressure dependence of collision distance. C_v = C_p - R = 35.4 - 8.31 = 27.09 J mol⁻¹ K⁻¹.γ = (C_p)/(C_v) = (35.4)/(27.09) ≈ 1.31. Substituting values gives 1.31, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter

View Details