Upload
lytu
View
240
Download
8
Embed Size (px)
Citation preview
MME6701 Solution to Assignment #2
1. Obtain a relation needed to compute the change in Helmholtz free energy when the initial and final states are specified by their pressure and volume.
Here the function is F = F (P, V). Then the differential equation of the function is
𝑑𝐹 = 𝑀 𝑑𝑃 + 𝑁 𝑑𝑉
Since for V = V (T, P)
𝑑𝑉 = 𝑉𝛼 𝑑𝑇 − 𝑉𝛽 𝑑𝑃
we have
𝑑𝐹 = 𝑀𝑑𝑃 + 𝑁 𝑉𝛼𝑑𝑇 − 𝑉𝛽𝑑𝑃 = 𝑁𝑉𝛼𝑑𝑇 + 𝑀 − 𝑁𝑉𝛽 𝑑𝑃
Again, since for the function F = F (T, P),
𝑑𝐹 = −(𝑆 + 𝑃𝑉𝛼)𝑑𝑇 + 𝑃𝑉𝛽𝑑𝑃
Equating the coefficients of dT and dP from the above two equations, we get
𝑁𝑉𝛼 = −(𝑆 + 𝑃𝑉𝛼)
𝑀 − 𝑁𝑉𝛽 = 𝑃𝑉𝛽
From the first equation:
𝑁 = −𝑆 + 𝑃𝑉𝛼𝑉𝛼
= −𝑆𝑉𝛼
+ 𝑃
and from the second equation:
𝑀 = 𝑃𝑉𝛽 + 𝑁𝑉𝛽 = 𝑉𝛽 (𝑃 + 𝑁)
𝑀 = 𝑉𝛽 𝑃 − 𝑆 + 𝑃𝑉𝛼𝑉𝛼
= −𝑆𝛽𝑎
Then the desired functional relation is:
𝑑𝐹 = −𝑆𝛽𝑎
𝑑𝑃 −𝑆𝑉𝛼
+ 𝑃 𝑑𝑉
2. The initial state of one mole of a monatomic gas is 10 atm pressure and 300 K temperature. Calculate the change in entropy of the gas for (a) an isothermal decrease in pressure to 5 atm, (b) a reversible adiabatic expansion to a pressure of 5 atm, (c) a constant-volume decrease in pressure to 5 atm.
(a) Here the required function is: S = S (T, P). Then
𝑑𝑆 = 𝐶!𝑇𝑑𝑇 − 𝑉𝛼 𝑑𝑃
For isothermal process, dT = 0, and
𝑑𝑆! = − 𝑉𝛼 𝑑𝑃!
For ideal gas, V = nRT/P and α = 1/T. Then
𝑑𝑆 = − 𝑛𝑅𝑇𝑃
1𝑇𝑑𝑃 = − 𝑛𝑅
𝑑𝑃𝑃
Integrating between the limits of (S1, 10 atm) and (S2, 5 atm), we get
∆𝑆 = −(1 𝑚𝑜𝑙)(8.314 𝐽/𝑚𝑜𝑙 𝐾) ln5 𝑎𝑡𝑚10 𝑎𝑡𝑚
= 5.76 𝐽/𝑚𝑜𝑙 𝐾
(b) For adiabatic process, δQ = TdS = 0. Thus
∆𝑆 = 0
(c) Here the function is: S = S (V, P). The differential equation of this function is
𝑑𝑆 = 𝐶!𝑇𝑉𝛼
𝑑𝑉 + 𝛽𝐶!𝑇𝛼
𝑑𝑃
At constant volume, dV = 0, and
𝑑𝑆! = 𝛽𝐶!𝑇𝛼
𝑑𝑃!
For ideal monatomic gas, β = 1/P, CV = 3R/2, and α = 1/T. Then
𝑑𝑆 = 3𝑅2
𝑑𝑃𝑃
Integrating between the limits of (S1, 10 atm) and (S2, 5 atm), we get
∆𝑆 = 3(8.314 𝐽/𝑚𝑜𝑙 𝐾)
2ln
5 𝑎𝑡𝑚10 𝑎𝑡𝑚
= −8.64 𝐽/𝑚𝑜𝑙 𝐾
3. One mole of N2 gas is contained at 273 K and a pressure of 1 atm. The addition of 3000 joules of heat to the gas at constant pressure causes 832 joules of work to be done during the expansion. Calculate (a) the final state of the gas, (b) the values of ΔU and ΔH for the change of state, and (c) the values of cP and cV for N2. Assume that nitrogen behaves ideally and the above change of state is conducted reversibly.
The initial volume
𝑉! = 𝑛𝑅𝑇!𝑃!
= (1 𝑚𝑜𝑙) (0.082 𝑙 𝑎𝑡𝑚/𝑚𝑜𝑙 𝐾) (273 𝐾)
1 𝑎𝑡𝑚 = 22.34 𝑙
At constant pressure, the work done on the system
𝑊! = −𝑃(𝑉! − 𝑉!)
−(832 𝐽) 0.082 𝑙 𝑎𝑡𝑚8.314 𝐽
= −(1 𝑎𝑡𝑚) (𝑉! − 22.34 𝑙)
𝑉! = 30.55 𝑙
Then the final temperature
𝑇! = 𝑃! 𝑉!𝑛𝑅
= (1 𝑎𝑡𝑚) (30.55 𝑙)
(1 𝑚𝑜𝑙) (0.082 𝑙 𝑎𝑡𝑚/𝑚𝑜𝑙 𝐾) = 372.51 𝐾
Hence, the final state of the system is: P2 = 1 atm, V2 = 30.55 l, T2 = 372.51 K.
The internal energy of the system
∆𝑈 = 𝑄 + 𝑊 = 3000 − 832 J = 2168 J
Since for the ideal gas, dU = CV dT, then
𝐶! 𝑇! − 𝑇! = 2168 J
𝐶! 372.51 − 273 𝐾 = 2168 J
𝐶! = 21.79 J/K
And heat capacity at constant pressure
𝐶! = 𝐶! + 𝑅 = 21.79 + 8.314 = 30.104 J/K
Then, the change in enthalpy of the system is
∆𝐻 = 𝐶! 𝑇! − 𝑇! = 30.104 372.51 − 273 𝐾 = 2995.25 J
4. Copper exists in the state of 298 K temperature and 1 atm pressure. Calculate the temperature to which the copper must be raised at 1 atm pressure to cause the same increase in molar enthalpy as is caused by increasing its pressure to 1000 atm at 298 K. The molar volume of copper at 298 K is 7.09 cc, and the thermal expansion is 0.493x10–3 K–1. These values can be taken as being independent of pressure in the range 1–1000 atm.
For the function H = H (T, P)
𝑑𝐻 = 𝐶! 𝑑𝑇 + 𝑉 (1 − 𝑇𝛼) 𝑑𝑃
At constant temperature
𝑑𝐻! = 𝑉 (1 − 𝑇𝛼) 𝑑𝑃!
Considering V of solid does not vary too much with P, integrating
∆𝐻! = 𝑉 (1 − 𝑇𝛼) (𝑃! − 𝑃!)
= 7.09 𝑐𝑐 1 − 298 𝑥 0.493𝑥10!! 1000 − 1 𝑎𝑡𝑚
= −6042.33 𝑐𝑐 𝑎𝑡𝑚 𝑥 8.314 𝐽
82.06 𝑐𝑐 𝑎𝑡𝑚 = −612.19 𝐽
The same amount of enthalpy change is produced when the system undergoes a constant pressure process. Thus
∆𝐻! = 𝐶! 𝑑𝑇!
!!
−612.19 𝐽 = 22.6 + 5.6𝑥10!! 𝑇 𝑑𝑇!
!"#
−612.19 = 22.6 𝑇 + 2.8𝑥10!! 𝑇! − 22.6 (298) − 2.8𝑥10!! (298)
2.8𝑥10!! 𝑇! + 22.6 𝑇 + 6371.26 = 0
𝑇 = −22.6 ± 22.6 ! − 4 (2.8𝑥10!!)(6371.26)
2 (2.8𝑥10!!) = 292.51 𝐾
Hence the temperature is 292.51 K.