teacher.buet.ac.bdteacher.buet.ac.bd/bazlurrashid/mme6701/sol_assign_2.pdfTitle Microsoft Word -...
Transcript of teacher.buet.ac.bdteacher.buet.ac.bd/bazlurrashid/mme6701/sol_assign_2.pdfTitle Microsoft Word -...
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
๐๐น = ๐ ๐๐ + ๐ ๐๐
Since for V = V (T, P)
๐๐ = ๐๐ผ ๐๐ โ ๐๐ฝ ๐๐
we have
๐๐น = ๐๐๐ + ๐ ๐๐ผ๐๐ โ ๐๐ฝ๐๐ = ๐๐๐ผ๐๐ + ๐ โ ๐๐๐ฝ ๐๐
Again, since for the function F = F (T, P),
๐๐น = โ(๐ + ๐๐๐ผ)๐๐ + ๐๐๐ฝ๐๐
Equating the coefficients of dT and dP from the above two equations, we get
๐๐๐ผ = โ(๐ + ๐๐๐ผ)
๐ โ ๐๐๐ฝ = ๐๐๐ฝ
From the first equation:
๐ = โ๐ + ๐๐๐ผ๐๐ผ
= โ๐๐๐ผ
+ ๐
and from the second equation:
๐ = ๐๐๐ฝ + ๐๐๐ฝ = ๐๐ฝ (๐ + ๐)
๐ = ๐๐ฝ ๐ โ ๐ + ๐๐๐ผ๐๐ผ
= โ๐๐ฝ๐
Then the desired functional relation is:
๐๐น = โ๐๐ฝ๐
๐๐ โ๐๐๐ผ
+ ๐ ๐๐
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
๐๐ = ๐ถ!๐๐๐ โ ๐๐ผ ๐๐
For isothermal process, dT = 0, and
๐๐! = โ ๐๐ผ ๐๐!
For ideal gas, V = nRT/P and ฮฑ = 1/T. Then
๐๐ = โ ๐๐ ๐๐
1๐๐๐ = โ ๐๐
๐๐๐
Integrating between the limits of (S1, 10 atm) and (S2, 5 atm), we get
โ๐ = โ(1 ๐๐๐)(8.314 ๐ฝ/๐๐๐ ๐พ) ln5 ๐๐ก๐10 ๐๐ก๐
= 5.76 ๐ฝ/๐๐๐ ๐พ
(b) For adiabatic process, ฮดQ = TdS = 0. Thus
โ๐ = 0
(c) Here the function is: S = S (V, P). The differential equation of this function is
๐๐ = ๐ถ!๐๐๐ผ
๐๐ + ๐ฝ๐ถ!๐๐ผ
๐๐
At constant volume, dV = 0, and
๐๐! = ๐ฝ๐ถ!๐๐ผ
๐๐!
For ideal monatomic gas, ฮฒ = 1/P, CV = 3R/2, and ฮฑ = 1/T. Then
๐๐ = 3๐ 2
๐๐๐
Integrating between the limits of (S1, 10 atm) and (S2, 5 atm), we get
โ๐ = 3(8.314 ๐ฝ/๐๐๐ ๐พ)
2ln
5 ๐๐ก๐10 ๐๐ก๐
= โ8.64 ๐ฝ/๐๐๐ ๐พ
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 ฮU and ฮH 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
๐! = ๐๐ ๐!๐!
= (1 ๐๐๐) (0.082 ๐ ๐๐ก๐/๐๐๐ ๐พ) (273 ๐พ)
1 ๐๐ก๐ = 22.34 ๐
At constant pressure, the work done on the system
๐! = โ๐(๐! โ ๐!)
โ(832 ๐ฝ) 0.082 ๐ ๐๐ก๐8.314 ๐ฝ
= โ(1 ๐๐ก๐) (๐! โ 22.34 ๐)
๐! = 30.55 ๐
Then the final temperature
๐! = ๐! ๐!๐๐
= (1 ๐๐ก๐) (30.55 ๐)
(1 ๐๐๐) (0.082 ๐ ๐๐ก๐/๐๐๐ ๐พ) = 372.51 ๐พ
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
โ๐ = ๐ + ๐ = 3000 โ 832 J = 2168 J
Since for the ideal gas, dU = CV dT, then
๐ถ! ๐! โ ๐! = 2168 J
๐ถ! 372.51 โ 273 ๐พ = 2168 J
๐ถ! = 21.79 J/K
And heat capacity at constant pressure
๐ถ! = ๐ถ! + ๐ = 21.79 + 8.314 = 30.104 J/K
Then, the change in enthalpy of the system is
โ๐ป = ๐ถ! ๐! โ ๐! = 30.104 372.51 โ 273 ๐พ = 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 0.493x10โ3 Kโ1. These values can be taken as being independent of pressure in the range 1โ1000 atm.
For the function H = H (T, P)
๐๐ป = ๐ถ! ๐๐ + ๐ (1 โ ๐๐ผ) ๐๐
At constant temperature
๐๐ป! = ๐ (1 โ ๐๐ผ) ๐๐!
Considering V of solid does not vary too much with P, integrating
โ๐ป! = ๐ (1 โ ๐๐ผ) (๐! โ ๐!)
= 7.09 ๐๐ 1 โ 298 ๐ฅ 0.493๐ฅ10!! 1000 โ 1 ๐๐ก๐
= โ6042.33 ๐๐ ๐๐ก๐ ๐ฅ 8.314 ๐ฝ
82.06 ๐๐ ๐๐ก๐ = โ612.19 ๐ฝ
The same amount of enthalpy change is produced when the system undergoes a constant pressure process. Thus
โ๐ป! = ๐ถ! ๐๐!
!!
โ612.19 ๐ฝ = 22.6 + 5.6๐ฅ10!! ๐ ๐๐!
!"#
โ612.19 = 22.6 ๐ + 2.8๐ฅ10!! ๐! โ 22.6 (298) โ 2.8๐ฅ10!! (298)
2.8๐ฅ10!! ๐! + 22.6 ๐ + 6371.26 = 0
๐ = โ22.6 ยฑ 22.6 ! โ 4 (2.8๐ฅ10!!)(6371.26)
2 (2.8๐ฅ10!!) = 292.51 ๐พ
Hence the temperature is 292.51 K.