Fluids, Gases, Thermodynamics, solutions

Problem 1:  (D)

Equation of continuity and Bernoulli's equation

v1A1 = v2A2,  P1 + ρgh1 + ½ρv12 = P2 + ρgh2 + ½ρv22.
v2 = 4v1,  h2 = h1, P2 - P1 = ½ρ(v22 - v12) = ½(1000 kg/m3)(15 m2/s2) = 7500 N/m2.

Problem 2:  (A)

Maxwell speed distribution

dn(v)/dv = 0,  2Avexp(-mv2/(2kT)) - (mv/(kT))Av2exp(-mv2/(2kT)) = 0,  v2 = 2kT/m.

Problem 3:  (E)

The ideal gas law, P = nRT/V

W = ∫V1V2 PdV = ∫V1V2 (RT/V) dV = RT ln(V2/V1). 

Problem 4:  (D)

Latent heat of fusion, ΔQ = m*Lf

Heat of fusion = power *time/mass = (100 V)*(10 A)*(1020 s)/(3 kg) ~ 3.4*105 J/kg.

Problem 5:  (C)

Stefan-Boltzmann law, radiated power = emissivity * σ * T4 * Area

Energy received is equal to energy emitted per unit time.
constant_1*πr2/(4πR2) = constant_2*4πr2*T4.
T is proportional to 1/R½.

Problem 6:  (C)

Specific heat,  ΔQ  = cmΔT

ΔQ = ΔU at constant volume.  U = (3/2)kT,    ΔQ/(mΔT) = (3/2)k/m.
Specific heat per atom:  (3/2)k.

Problem 7:  (B)

The PV diagram, dW = PdV

W = ½(VC - VA)(PB - PC) = area under the curve.

Problem 8:  (E)

Entropy

dS = dQ/T = c m dT/T,  ΔS = c m ln(T2/T1).

Problem 9:  (D)

Entropy

dS = -dQ/T1 + dQ/T2.

Problem 10:  (C)

Entropy

The entropy of the gas in process (3) decreases and therefore the entropy of the reservoir (4) increases, since the total entropy for a reversible process does not change.  For an  isothermal process the internal energy of a gas is only a function of its temperature, ΔU = 0.  For the isothermal compression of an ideal gas we have for the work done by the gas.

W = ∫V1V2 PdV = ∫V1V2 (nRT/V) dV = nRT ln(V2/V1). 

W is negative if V2 < V1.  Since ΔU = 0, the heat transferred to the gas is ΔQ = W.  Therefore for one mole of gas ΔQ = RT ln(V2/V1) and ΔS = R ln(V2/V1).