Chapter 3, Exercises #3.4(b), 3.11(b), 3.12(b), 3.13(b); Problems #3.1, 3.2

Chapter 4, Exercises #4.1(b), 4.3(b), 4.5(b), 4.6(b)
 

3.4(b)
    
 

3.11(b)
    
 
 
 
 
 
 
 

3.12(b)

V = V[0.77 + 3.7x10^-4T + 1.52x10^-6T²]

= (1/V)(V/T)_p = (1/V)(V[3.7x10^-4 + 3.04x10^-6T]
 

=[3.7x10^-4 +3.04x10^-6T]/[0.77 + 3.7x10^-4T + 1.52x10^-6T²] = 1.273x10^-3/K at 310K
 

3.13(b)

= -(1/V)(V/p)_T = 2.21x10^-6 /atm
 

= m/V (/p)_T = (-m/V²)(V/p)_T from the chain rule and
 

= (m/V)(-1/V)(V/p)_T =
 

Thus d/ = dp and integrating, n(₂/₁) = (p₂ - p₁) = p

For the density to increase by 8% implies that ₂ = 1.08₁
 

p = n(1.08)/2.21x10^-6 = 3.48x10⁴ atm p₂ (That is, p₁ is only 1 atm.)
 

3.1

= 2.21x10^-6 atm^-1 (1atm/1.01x10⁵Pa) = 2.19x10^-11Pa^-1
 

at 1.000 km, p = gh = (1.03g/cm³)(1m³/10⁶cm³)(10^-3kg/1g)(9.81m/s²)(1000m)

= 1.01x10⁷Pa
 

dV/V = -dp at constant T. Thus, n(V₂/V₁) = -p, and V₂ = V₁exp(-p).
 

V₂ = (10³cm³)exp(-2.19x10^-11Pa^-1 x 1.01x10⁷Pa) = 999.8cm³
 

Since dV = VdT - Vdp, we can fix T and change p; and then fix p and change T, since V is a state function:
 

(Assuming that and are constants) n(V₂/V₁) = T and V₂ = V₁exp(T)
 

Thus, the new V₂ = 999.8exp(8.61x10^-5K^-1 x 30K) = 997.2cm³
 
 
 

3.2

There are several ways to approach this problem depending on the data available. For example, for this process at constant p: dH = C_pdT and dU = C_VdT + _TdV. Since dV = VdT - Vdp = VdT at constant p, dU = (C_V + _TV)dT. Or, one can find U from U = H -(pV).

Using the second method, dH = C_pdT or H = C_pT for constant C_p = 75.291J/mol-K from Table 2.6. Thus H = 752.91 for a 10K increase in T.
 

U = H - p(V₂ - V₁) where p = 1 bar = 10⁵Pa.

Since dV = VdT at constant pressure or n(V₂/V₁) = T, then

(V₂ -V₁) = V₁(exp(T) - 1) = 18cm³(exp(2.1x10^-4(10)) - 1) = 18(2.1x10^-3)
 

and U = 752.91J - 10⁵Pa(18cm³)(1m³/10⁶cm³)(2.1x10^-3) = 752.91-0.003 = 752.91J
 

The small difference in energy and enthalpy for this small temperature increase is due to the work that is done by thermal expansion of the water as it is warmed by 10 degrees.
 

4.1(b)

dS = dq_rev/T
 

a) S = 50kJ/273K = 183J/K

b) S = 50kJ/343K = 146J/K
 

4.3(b)

per mole C_V = 5/2R. For an ideal (perfect), gas per C_p = C_V + R = 7/2R

At constant p, dq_rev = dH

S = ₂₇₃³⁷³[C_p/T]dT = 7/2Rn(373/273) = 9.08J/K
 

4.5(b)

C_V = 27.5J/K-mol q = 0 for adiabatic process
 

S = 0 for reversible adiabatic process
 

dU = C_VdT = dw for the ideal gas
 

U = C_VT = w for constant heat capacity, U = w = 2mol(27.5J/mol-K)(300K-250K)

U = w = 2.75kJ
 

H = U + (pV) = U + nRT = 2.75kJ + 2(8.31J/mol-K)(50K) = 3.58kJ
 

4.6(b)

C_p = 5/2R, T₁ = 250K, T₂ = 700K, V₁ = 20.0L, V₂=60.0L, ideal gas
 

GAS(T₁,V₁) GAS(T₂, V₁) GAS(T₂, V₂)
 

For step 1, constant volume heating: dq_rev = dU = C_VdT = (C_p -R)dT = 3/2R
 

S₁ = [C_V/T]dT = C_Vn(T₂/T₁) = (3.5mol)(3/2R)n(700/250) = 44.9J/K
 

For step 2, isothermal expansion: dU = C_VdT + _TdV = 0 for ideal gas

Thus dq = -dw and for reversible expansion dq_rev = +pdV = +nRT₂dV/V
 

Thus, dS = dq_rev/T₂ = nRdV/V or S = nRn(V₂/V₁) = 3.5mol(8.31J/mol-K)n(60.0/20.0)

S₂ = 32.0J/K and S = S₁ + S₂ = 44.9 + 32.0 = 76.9J/K