While the pressure remains constant at 689.5 kpaa the volume of a system of air changes from 0.567 m3 to 0.2283 m3. Determine (a) ΔU, (b) ΔH, (c) Q, (d) ΔS. (e) If the process is nonflow and internally reversible, what is the work? Cp = 0.24Btu/(lbm.° R), Cv = 0.1714Btu/(lbm.° R), R= 53.34(ft-lbf)/(lbm.° R)
To solve this problem, we can utilize the first law of thermodynamics, which states that:
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat transfer, and W is the work done.
We are given the initial and final volumes of the system, as well as the constant pressure. We can calculate the work done using the equation:
W = PΔV
where P is the pressure and ΔV is the change in volume.
(a) To find ΔU, we can use the ideal gas equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
We can assume that the number of moles (n) and the gas constant (R) are constant since they are not given to change.
Using the ideal gas equation:
P1V1 = nRT1
P2V2 = nRT2
Dividing the second equation by the first equation:
(P2V2) / (P1V1) = (nRT2) / (nRT1)
Simplifying:
(V2 / V1) = (T2) / (T1)
Plugging in the given values:
(V2 / 0.567 m³) = (T2) / (T1)
Solving for T2:
T2 = (V2 / 0.567 m³) * T1
Plugging in the values:
T2 = (0.2283 m³ / 0.567 m³) * T1
T2 = 0.4026 * T1
Now we know the initial and final temperatures, we can use the specific heat capacities (Cp and Cv) to find the change in internal energy ΔU:
ΔU = n * (Cv * ΔT)
ΔU = n * (Cv * (T2 - T1))
Plugging in the values:
ΔU = n * (0.1714 Btu/(lbm.°R) * (0.4026 * T1 - T1))
(b) To find ΔH, we can use the equation:
ΔH = ΔU + Δ(PV)
Since the pressure is constant, Δ(PV) = PΔV
ΔH = ΔU + PΔV
Plugging in the values:
ΔH = n * (0.1714 Btu/(lbm.°R) * (0.4026 * T1 - T1)) + P * ΔV
(c) To find Q, we can use the equation:
Q = ΔH - ΔPV
Plugging in the values:
Q = n * (0.1714 Btu/(lbm.°R) * (0.4026 * T1 - T1)) + P * (V2 - V1)
(d) To find ΔS, we can use the equation:
ΔS = n * Cv * ln(T2 / T1)
Plugging in the values:
ΔS = n * 0.1714 Btu/(lbm.°R) * ln((0.4026 * T1) / T1)
(e) If the process is nonflow and internally reversible, the work done is given by:
W = -ΔU
Plugging in the value of ΔU, we calculated earlier.
To determine the values of ΔU (change in internal energy), ΔH (change in enthalpy), Q (heat transfer), and ΔS (change in entropy), we can use the first law of thermodynamics:
ΔU = Q - W (Equation 1)
ΔH = ΔU + PΔV (Equation 2)
Q = ΔU + W (Equation 3)
ΔS = Q/T (Equation 4)
Let's calculate each value step by step:
(a) To calculate ΔU, we need to know the initial and final internal energy, U₁ and U₂, respectively, for a system of air.
Since we are given constant pressure, we can use the equation:
ΔU = Cv * (T₂ - T₁) (Equation 5)
Here, Cv represents the specific heat at constant volume, and T₁ and T₂ are the initial and final temperatures. To calculate the temperatures, we can use the ideal gas law:
P₁V₁/T₁ = P₂V₂/T₂ (Equation 6)
Given that the pressure P remains constant at 689.5 kPa, we can rewrite Equation 6 as:
V₁/T₁ = V₂/T₂ (Equation 7)
Rearranging Equation 7 to solve for T₂:
T₂ = (V₂ * T₁) / V₁ (Equation 8)
Now we can substitute Equation 8 into Equation 5:
ΔU = Cv * [(V₂ * T₁) / V₁ - T₁] (Equation 9)
Using the provided values, we can substitute them into Equation 9 to calculate ΔU.
(b) To calculate ΔH, we can use Equation 2. We already calculated ΔU in step (a), and P represents the constant pressure. So, we have:
ΔH = ΔU + P * (V₂ - V₁) (Equation 10)
Substitute the values into Equation 10 to calculate ΔH.
(c) To calculate Q, we can use Equation 3. Again, we already have ΔU from step (a), and we need to calculate W (work done by the system).
However, since the process is nonflow and internally reversible, we know that there is no work done (W = 0). Thus, the equation simplifies to:
Q = ΔU (Equation 11)
Substitute the ΔU value into Equation 11 to calculate Q.
(d) To calculate ΔS, we use Equation 4. We already have Q from step (c), and T represents the temperature.
ΔS = Q / T (Equation 12)
Substitute the values into Equation 12 to calculate ΔS.
(e) Finally, if the process is nonflow and internally reversible, the work (W) is equal to zero.