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#temperature calculation

12 public questions tagged with this topic.

Wheatstone Bridge and Meter Bridge

A nichrome wire has a resistance of \( 50 \, \Omega \) at \( 25^\circ \text{C} \) and \( 55 \, \Omega \) at a higher tem

**Unbalanced bridge** has potential difference between galvanometer nodes, current direction determined by which node higher potential, i.e., if R₁/R₂ > R₃/R₄, left node higher, current flows one way, else opposite. Galvanometer deflection indicates imbalance magnitude. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: 55 = 50 [1 + 1.7 × 10⁻⁴ (T - 25)] . Solve: 55 = 50 + 50 × 1.7 × 10⁻⁴ (T - 25) ⇒ 5 = 8.5 × 10⁻³ (T - 25) . T - 25 = (5/8.5 × 10⁻³) ≈ 588 ⇒ T ≈ 613° C . Applying I = n e A v_d, R

Ref: NCERT > Physics Book > Current Electricity > Wheatstone Bridge and Meter Bridge

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Second Law Heat Engines and Kelvin-Planck

0.4 moles of an ideal gas at 340 K expand adiabatically from 7 atm to 1 atm. If gamma = 1.4 , what is the final temperat

**Carnot engine** reversible engine operating between T_h and T_c has maximum efficiency η =1 - T_c/T_h, T in kelvin, e.g., T_h=400 K T_c=300 K η=0.25, real engines less due to irreversibilities, second law defines direction of spontaneous processes and entropy increase. Adiabatic: T₁ V₁^γ-1 = T₂ V₂^γ-1 , V₁ = (μ R T₁)/(P₁) = (0.4 × 8.3 × 340)/(7) ≈ 161.37 L , V₂ = (0.4 × 8.3 × T₂)/(1) = 3.32 T₂ . 340 × 161.37⁰.4 = T₂ × (3.32 T₂)⁰.4 . Approximate: T₂ ≈ 245 K . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W =

Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck

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Partial Pressures and Gas Mixtures

The rms speed of a gas is 400 m/s at 100 K. At what temperature will the rms speed be 800 m/s?

**Partial pressure concept** for mixture of non-reacting gases, total pressure sum of partials, each gas behaves independently, kinetic theory still holds with effective n = Σ n_i, mean free path depends on total n and cross-sections, mixture properties weighted by mole tions. v_rms ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(800)/(400) = √((T₂)/(100)), 2 = √((T₂)/(100)).Square both sides: 4 = (T₂)/(100), T₂ = 400 K. Substituting values gives 400 K, 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 > Partial Pressures and Gas Mixtures

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RMS Speed and Temperature Dependence

The rms speed of a gas is 450 m/s at 225 K. At what temperature will the rms speed be 900 m/s?

**Temperature dependence of RMS speed** v_rms ∝ √T, doubling T increases v_rms by √2≈1.414, e.g., at 300 K v_rms for O₂ ≈483 m/s, at 600 K ≈683 m/s, illustrating kinetic theory relation between temperature and molecular motion, average kinetic energy ½ m v_rms² =3/2 k_B T. v_rms ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(900)/(450) = √((T₂)/(225)), 2 = √((T₂)/(225)).Square both sides: 4 = (T₂)/(225), T₂ = 900 K. Substituting values gives 900 K, 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 > RMS Speed and Temperature Dependence

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RMS Speed and Temperature Dependence

At what temperature is the rms speed of oxygen molecules 600 m/s? (Molecular mass of O₂ = 32 u, k_B = 1.38 × 10⁻²³ J K⁻¹

**RMS speed** v_rms = √(3 R T/M) = √(3 k_B T/m) where M molar mass (kg/mol), m molecular mass (kg), k_B=1.38×10⁻²/³ J/K, R=8.314 J/mol·K, T absolute temperature (K). Proportional to √T and 1/√M, lighter gases faster at same T, e.g., H₂ faster than O₂, temperature increase raises v_rms as √T. v_rms = √((3k_B T)/(m)), m = 32 × 10⁻³⁶.02 × 10²³ = 5.32 × 10⁻²⁶ kg.600² = 3 × 1.38 × 10⁻²/³ × T5.32 × 10⁻²⁶, T = 3.6 × 10⁵ × 5.32 × 10⁻²⁶/⁴.14 × 10⁻²/³ ≈ 463 K. Substituting values gives 463 K, which matches expected kinetic theory result, confirming mean free path

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > RMS Speed and Temperature Dependence

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RMS Speed and Temperature Dependence

At what temperature is the rms speed of hydrogen molecules 2000 m/s? (Molecular mass of H₂ = 2 u, k_B = 1.38 × 10⁻²³ J K

**RMS speed** v_rms = √(3 R T/M) = √(3 k_B T/m) where M molar mass (kg/mol), m molecular mass (kg), k_B=1.38×10⁻²/³ J/K, R=8.314 J/mol·K, T absolute temperature (K). Proportional to √T and 1/√M, lighter gases faster at same T, e.g., H₂ faster than O₂, temperature increase raises v_rms as √T. v_rms = √((3k_B T)/(m)), m = 2 × 10⁻³⁶.02 × 10²³ = 3.32 × 10⁻²⁷ kg.2000² = 3 × 1.38 × 10⁻²/³ × T3.32 × 10⁻²⁷, T = 4 × 10⁶ × 3.32 × 10⁻²⁷/⁴.14 × 10⁻²/³ ≈ 321 K. Substituting values gives 321 K, which matches expected kinetic theory result, confirming mean free path

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > RMS Speed and Temperature Dependence

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Degrees of Freedom and Molar Specific Heat

At what temperature is the rms speed of oxygen molecules 964 m/s? (Molecular mass of O₂ = 32 u, k_B = 1.38 × 10⁻²³ J K⁻¹

**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. v_rms = √((3k_B T)/(m)), m = 32 × 10⁻³⁶.02 × 10²³ = 5.32 × 10⁻²⁶ kg.964² = 3 × 1.38 × 10⁻²/³ × T5.32 × 10⁻²⁶, T = 9.29 × 10⁵ × 5.32 × 10⁻²⁶/⁴.14 × 10⁻²/³ ≈ 1194 K. Substituting values gives 1200 K, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n

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

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Degrees of Freedom and Molar Specific Heat

The rms speed of a gas is 300 m/s at 150 K. At what temperature will the rms speed be 600 m/s?

**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. v_rms ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(600)/(300) = √((T₂)/(150)), 2 = √((T₂)/(150)).Square both sides: 4 = (T₂)/(150), T₂ = 600 K. Substituting values gives 600 K, 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

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Gas Laws and Volume-Temperature Relations

At what temperature is the rms speed of helium molecules 1500 m/s? (Atomic mass of He = 4 u, k_B = 1.38 × 10⁻²³ J K⁻¹)

**Gas laws** Boyle, Charles, Gay-Lussac are special cases of ideal gas equation, for constant pressure volume-temperature relation V ∝ T, for constant temperature pressure-volume inverse, for constant volume pressure-temperature direct, enabling calculation of new volume from temperature ratio. v_rms = √((3k_B T)/(m)), m = 4 × 10⁻³⁶.02 × 10²³ = 6.64 × 10⁻²⁷ kg.1500² = 3 × 1.38 × 10⁻²/³ × T6.64 × 10⁻²⁷, T = 2.25 × 10⁶ × 6.64 × 10⁻²⁷/⁴.14 × 10⁻²/³ ≈ 361 K. Substituting values gives 361 K, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

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Mean Free Path and Molecular Diameter

The rms speed of a gas is 300 m/s at 150 K. At what temperature will the rms speed be 600 m/s?

**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. v_rms ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(600)/(300) = √((T₂)/(150)), 2 = √((T₂)/(150)).Square both sides: 4 = (T₂)/(150), T₂ = 600 K. Substituting values gives 600 K, 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

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