Problem 1:

(a)  When a gas isothermally expands against a fixed external pressure, would the expansion be reversible or irreversible?  Explain your answer.
(b)  Will the entropy of an ideal gas increase, decrease of remain fixed during the expansion?  Explain your answer.
(c)  Suppose the external pressure is zero, what would be the entropy change of the gas, its surroundings, and for the universe?  If appropriate, express your answer in terms of temperature and volume.

Solution:

• Concepts:
  The ideal gas law, isothermal processes, the first law of thermodynamics
• Reasoning:
  PV = Nk_BT,  dU = dQ - dW,
  Increase in internal energy of a system
  = heat put into the system - work done by the system on its surroundings,
• Details of the calculation:
  (a)  An expansion is reversible only when the pressure of the gas inside and outside are the same during all stages of the expansion.  Only in this case could an infinitesimal change in pressure reverse the direction of the expansion (i.e., resulting in compression).  This is NOT the case when the outside pressure is constant, because when the pressures are different, one cannot reverse the direction of change via an infinitesimal change of the inside or outside pressure.
  
  (b)  The entropy of the gas depends on temperature and volume.  The temperature is fixed but the volume increases.  The latter increases the multiplicity of the gas and therefore its entropy increases.
  
  (c)   To calculate the entropy change of the gas, we need to imagine a reversible path.  The easiest one to calculate is to assume a reversible isothermal expansion.  Accordingly
  dU = 0 (constant temperature),
  dQ = dW = P_gasdV = (Nk_BT/V)dV.
  or
  Q = Nk_BT ln⁡(V_f/V_i)  and ∆S = Nk_Bln⁡(V_f/V_i).         
  Since entropy is a state function, this equation also applies to an irreversible isothermal expansion.
  
  Now, since the outside pressure is zero and the process is isothermal, we have dU = 0 and dW = 0.  Hence dQ = 0, meaning there is no heat exchange with the surroundings (∆S_surr = 0). We then find the following answer:
  ∆S_universe = ∆S_system + ∆S_surr = Nk_Bln(V_f/V_i).

Problem 2:

A glass is filled to height h₀ with a volume of water V₀ with density ρ.  A straw with uniform cross sectional area A is used to drink the water by creating a pressure at the top of the straw (P_top) that is less than atmospheric pressure (P_Atm).  The top of the straw is a distance h_top above the bottom of the glass.

(a)  Determine the pressure (P_top) at the top of the straw as a function of the height (h) of the water in the glass so that the velocity of water coming out of the top of the straw remains constant at a value of v_top.
(b)  How much work is done by the person in order to drink all the water in the glass with the constant velocity v_top?
(c)  How much time does it take to drink all the water in the glass with the constant velocity v_top?

Solution:

• Concepts:
  Bernoulli's equation, work and energy
• Reasoning:
  Bernoulli's equation implies energy conservation.  We ignore viscosity.
• Details of the calculation:
  Let the y axis point up and the bottom of the glass be at y = 0.
  (a)  P_atm + ρgh + ½ρv² = P_top + ρgh_top + ½ρv_top².
  Here h is the height of the water in the glass and v is the speed with which h decreases.
  Assume that v is negligible compared to v_top.  Then
  P_top = P_atm - ρg(h_top - h) - ½ρv_top².
  
  (b)  The water has to be lifted and given some kinetic energy.
  The work done lifting the water is
  W = ∫₀^h0A_glassρdh g(h_top - h) = A_glassρgh₀(h_top - h₀/2) = V₀ρg(h_top - h₀/2), since A_glassh₀ = V₀.
  W_kin = ½V₀ρv_top².  W_total = V₀ρ[g(h_top - h₀/2) + ½v_top²].
  
  (c)  The volume flow rate is Av_top.  The time it takes to drink all the water is t = V₀/(Av_top).

Problem 3:

N atoms, each of mass m, of an ideal monatomic gas occupy a volume consisting of two identical chambers connected by a narrow tube, as illustrated on the right.  There is a gravitational field with acceleration g directed downward in the figure.  The gas is in thermal equilibrium at temperature T.

The height h of the upper chamber above the lower chamber is much greater than the height l of either chamber.  The volume of the tube is negligible compared with that of the chambers. 
Assume that mgl << k_BT, but not that there is any particular relation between mgh and k_BT.  Treat the system classically and assume the atoms as having no internal degrees of freedom.
(a)  Calculate the number of atoms in the upper chamber in terms of N, m, g, h, and T.
(b)  Calculate the total energy (kinetic and potential) of the system in terms of the same parameters.
(c)  From the total energy obtain an expression for the specific heat of the system as a function of the temperature.

Solution:

• Concepts:
  Boltzmann distribution
• Reasoning:
  The probability of finding an atom at height z is proportional to exp(-mgz/(k_BT)).
• Details of the calculation:
  (a)  N_upper/N_lower = exp(-mgh/(k_BT)).  N_upper + N_lower = N.
  N_upper(exp(mgh/(k_BT) + 1) = N.  N_upper = N/(exp(mgh/(k_BT) + 1).
  
  (b)  E_kin = (3/2)Nk_BT,  E_pot = N_uppermgh = Nmgh/(exp(mgh/(k_BT) + 1). 
  E_tot = N((3/2)k_BT + mgh/(exp(mgh/(k_BT) + 1)).
  (We set E_pot = 0 at z = 0.)
  
  (c)  Specific heat per atom c_V = (1/N)dQ/dT = (1/N)dE_tot/dT.
  c_V = (3/2)k_B + k_B(mgh/(k_BT))² exp(mgh/(k_BT)/(exp(mgh/(k_BT) + 1)².
  Specific heat of the system = Nc_V.

Problem 4:

(a)  One mole of ideal gas with constant heat capacity C_V is placed inside a cylinder.  Inside the cylinder there is a piston which can move without friction along the vertical axis.  Pressure P₁ is applied to the piston and the gas temperature is T₁. 
At some point, P₁ is abruptly changed to P₂ (e.g. by adding or removing a weight from the piston).  As a result, the gas volume changes adiabatically.  Find the temperature T₂ and the volume V₂ after the thermodynamic equilibrium has been reached in terms of C_V, P₁, T₁, and P₂.  Use the relation between heat capacities C_V and C_P to simplify the formulas.
Definition of C_V:  dU = C_VdT, C_P = C_V + R

(b)  After the thermodynamic equilibrium has been established in part (a), the pressure is abruptly reset to its original value P₁.  Compute final values of the temperature T_f and the volume V_f after the thermodynamic equilibrium has been reached again. 
Compute the difference in temperatures (T_f  - T₁) and show that it is quadratic in (P₂ - P₁).
Comment on the sign of the temperature difference.

Solution:

• Concepts:
  The ideal gas law, adiabatic processes, the first law of thermodynamics
• Reasoning:
  ΔQ = 0 for adiabatic processes, and thus the first law of thermodynamics becomes
  ΔU + ΔW = 0, where ΔW is the work done by the gas, and U is its internal energy.
• Details of the calculation:
  (a)  Using ΔW = P₂ΔV and ΔU = C_VΔT we obtain:
  C_V(T₂ - T₁) + P₂(V₂ - V₁) = 0.
  
  Using the ideal gas PV = RT we get
  C_VT₂ = C_VT₁ - RT₂ + P₂V₁.
  T₂ = (C_VT₁ + P₂RT₁/P₁)/(C_V + R) = T₁(C_V + RP₂/P₁)/(C_V + R),
  V₂ = RT₂/P₂ = (C_VT₁ R/P₂ + R²T₁/P₁)/(C_V + R).
  Using the relation between heat capacities C_P - C_V = R we get
  T₂ = T₁(C_V + RP₂/P₁)/C_p,
  V₂ = RT₂/P₂ = (RT₁/P₁)(C_VP₁/P₂ + R)/C_p = V₁(C_VP₁/P₂ + R)/C_p.
  
  (b)   Using the result from part (a) we immediately obtain
  T_f = T₂(C_V + RP₁/P₂)/C_p,
  V_f = (RT₂/P₂)(C_VP₂/P₁ + R)/C_p = V₂(C_VP₂/P₁ + R)/C_p.
  T_f - T₁ = T₁(C_V + RP₂/P₁)(C_V + RP₁/P₂)/C_p² - T₁
  = T₁(C_V² + R² + C_VRP₂/P₁ + C_VRP₁/P₂)/C_p² - T₁
  = RT₁C_V (-2 + P₂/P₁ + P₁/P₂)/C_p²
  = (V₁/P₂)(C_V/C_p²)(-2P₁P₂ + P₂² + P₁²)
  = (V₁/P₂)(C_V/C_p²)(P₂ - P₁)²
  The change in temperature is quadratic in (P₂ - P₁).  It is always positive, as expected from the second law of thermodynamics.

Problem 5:

Two party balloons have a volume of 0.03 m³ each.  One is filled with air at 1.1 atmospheric pressure and its mass, including skin is 40.6 g.  The other is filled with helium at 1.1 atmospheric pressure and its mass, including skin, is 6.8 g.  The density of the air is 1.2 kg/m³.
A science museum has built an "elevator" which consists of a chamber with scales at the top and the bottom that can read the forces pushing against or pulling on the floor or the ceiling of the cart.  At the push of a button, the cart accelerates at a rate of 2 m/s² upward for 2 s, and then it decelerate at the same rate until it stops at its maximum height.
The air balloon is suspended from the ceiling with a string of negligible mass, and the helium balloon is fixed to the floor with the same type of string.
(a)  What is the maximum height reached by the bottom of the cart?
(b)  What is the tension in the strings when the cart is at rest, and in which direction do the strings pull on each balloon?
(c)  What is the tension in the strings while the car is accelerating upward, and in which direction do the strings pull on each balloon?

Let g = 10 m/s².  Assume the density of the air is constant in the chamber.

Solution:

• Concepts:
  Buoyancy
• Reasoning:
  The buoyant force is the force the surrounding fluid would have to exert on the displaced fluid to accomplish the state of motion given in the problem in the absence of the object.
• Details of the calculation:
  (a)  h = 2*½g(2s)² = 40 m.
  
  (b)  The cart at rest is an inertial frame.  The forces acting on each balloons are gravity, the buoyant force and the force due to the tension in the string.  Let the y-axis point upward.
  F_net = 0 = -m_balloong j + ρ_airV_balloong j  + T j.
  air balloon:  T j = (40.6 *10^-3*10 N - 1.2*0.03 m³*10 N) j = 0.046 N j.
  The tension is 0.046 N and the sting pulls the balloon up.
  helium balloon:  T j = (6.8 *10^-3*10 N - 1.2*0.03 m³*10 N) j = -0.29 N j.
  The tension is 0.29 N and the sting pulls the balloon down.
  
  (c)  We can choose the non-inertial reference frame of the accelerating car or the inertial reference frame of the ground.
  Choosing the inertial frame the forces acting on each balloons are gravity, the buoyant force and the force due to the tension in the string. 
  We have F_net = m_balloona j = -m_balloong j + ρ_airV_balloon(g + a) j  + T j.
  T j = m_balloon(a + g) j - ρ_airV_balloon(g + a) j.
  Air balloon accelerating upward:  T j = (40.6 *10^-3*12 N - 1.2*0.03 m³*12 n) j = 0.055 N j.
  The tension is 0.055 N and the sting pulls the balloon up.  The tension in the string increases.
  Helium balloon:  T j = (6.8 *10^-3*12 N - 1.2*0.03 m³*12 n) j = -0.35 N j.
  The tension is 0.35 N and the string pulls the balloon down.  The tension in the string increases.