More Problems

Problem 1:

On the basis of hydrostatic equilibrium, one can make an estimate that the pressure at the center of the Sun is 5*10¹⁴ N/m². 
(a)  If the Sun is purely hydrogen (with a number density of 3*10³⁰/m³) and can be treated as an ideal gas, what is the temperature at the solar center?  
(b)  What is the typical velocity of hydrogen ions in the solar center?

Solution:

• Concepts:
  Ideal gas law:  P = nk_BT 
  Kinetic theory:  U = ½mv² = (3/2)k_BT.
• Reasoning:
  We assume that is composed of purely hydrogen gas.
• Details of the calculation:
  (a)  T = P/(nk_B) = (5*10¹⁴ N/m²)/((1.38*10^-23 J/K)*(3*10³⁰/m³)) = 1.21*10⁷ K.
  (b)  v_rms = (3k_BT/m)^½ = 3*(1.38*10^-23 J/K)*(1.21*10⁷ K)/(1.67*10^-27 kg)^½ = 5.5*10⁵ m/s.

Problem 2:

Using the ideal gas model, find the specific heat at constant volume for Helium in units of J/(kg K).

Solution:

• Concepts:
  Specific heat, the ideal gas law
• Reasoning:
  For an ideal monatomic gas U = (3/2)NkT.
  dU/dT = dQ/dT at constant volume.
• Details of the calculation:
  dU/dT = (3/2)Nk, dQ/dT = Mc_V.
  c_V = 3Nk/(2M) = 3k/(2m), where m is the mass of a He atom.
  c_V = (3*1.38*10^-23 J/K)/(2*6.64*10^-27 kg) = 3120 J/(kg K).

Problem 3:

Sketch the entropy S of a 1 kg block of ice as a function of temperature T, as we take the ice from -10 ^oC to 110 ^oC.  Indicate where all phase changes occur.

Solution:

• Concepts:
  Entropy
• Reasoning:
  Change in entropy:   ΔS = ∫_i^f dS = ∫_i^f dQ_r/T
  Temperatures are absolute temperatures (K).
  The subscript r denotes a reversible path.
• Details of the calculation:
  Raising the temperature of the ice: 
   ΔS =  ∫_i^f dQ_r/T = mc_p∫_i^f dT/T = m c_{P_ice} ln(T_f/T_i).
  Melting the ice: ΔS = m L_fus/(273 K).
  Raising the temperature of the water:  ΔS = m c_{P_water} ln(T_f/T_i).
  Vaporizing the water: ΔS = m L_vap/(373 K) .
  Raising the temperature of the stem:  ΔS = m c_{P_steam} ln(T_f/T_i).
  
  ln(1 + ε) ≈ ε - ½ε² + ... .
  ln(T/T_i) = ln((T_i + ΔT )/T_i)  =  ln(1 + ΔT/T_i) ≈ ΔT/T_i.
  When raising the temperature of the ice, water, and steam in the given temperature range, the entropy increases approximately linearly with temperature.
  

Problem 4:

Joe is using a calibrated spectrometer with a diffraction grating with a line density of 1200 lines per mm to measure the temperature of a melted metal.  Two different wavelengths have their first-order maxima at 60 degrees and 45 degrees. The relative intensities of the two wavelengths are 10:1.  What is the temperature of the metal?

Solution:

• Concepts:
  Radiation laws
• Reasoning:
  For a grating: d sinθ = λ,  d = 1/1.2*10⁶,  λ₁ = d sin60^o = 722 nm,  λ₂ = d sin45^o = 589 nm.
  Plank radiation law:  I(λ,T) = (2hc²/λ⁵)[exp(hc/(kT λ)) - 1]^-1.
• Details of the calculation:
  Given:   I(λ₁,T)/I(λ₂,T) = 10 = (λ₂⁵/λ₁⁵)[exp(hc/(kT λ₂)) - 1]/[exp(hc/(kT λ₁)) - 1].
  hc/(λ₁k) = 1.99*10⁴ K,  hc/(λ₂k) = 2.44*10⁴ K. 
  Since I(722 nm) is greater than  I(589 nm) the temperature is below 5000 K, and
  exp(hc/(kTλ))  >>  1.  We therefore can write
  10 = (λ₂⁵/λ₁⁵)[exp(hc(λ₁ - λ₂)/(kT λ₂λ₁))].
  ln[10 (λ₁⁵/λ₂⁵)] = ln10 + 5 ln(1.22) = 3.32 = hc(λ₁ - λ₂)/(kT λ₂λ₁) = 4.48*10³/T.
  T = 1.35*10³ K.

Problem 5:

A child sitting in a train car holds a He-filled balloon with volume V = 1.4*10⁴ cm³ on a string.  The average density of the balloon is 1/5 the density of the surrounding air, ρ_air = 1.2 kg/m³.  The train is leaving the station, accelerating with constant acceleration a = 1 m/s² forward.  Find the force F (magnitude and direction) the string exerts on the balloon.

Solution:

• Concepts:
  Free-body diagrams, the buoyant force
• Reasoning:
  Let the x-axis point in the direction of a and let the y-axis point up.  Let ρ be the density of air.  For the net force F on the balloon we have F_y = 0, F_x = ma.
• Details of the calculation:
  The buoyant force acting on the balloon is B = ρVg j + ρVa i.
  (The buoyant force is the force the surrounding air would have to exert on the displaced airr to accomplish the state of motion given in the problem in the absence of the object.  The buoyant force has the same magnitude as the apparent weight of the displaced air and points in the opposite direction.)
  F_y - mg + B_y = 0.  F_y = ρVg/5 - ρVg = -4ρVg/5.
  F_x + B_x = ma.  F_x = ρVa/5 - ρVa = -4ρVa/5.
  |F| = (4ρV/5)(g² + a²)^1/2 = 0.132 N.  The force the string exerts on the balloon points down and backwards, making 180^o - θ with the y-axis.
  We have tanθ = a/g, θ = 5.8