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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 = + Since for V = V (T, P) = โˆ’ we have = + โˆ’ = + โˆ’ Again, since for the function F = F (T, P), = โˆ’( + ) + Equating the coefficients of dT and dP from the above two equations, we get = โˆ’( + ) โˆ’ = From the first equation: = โˆ’ + = โˆ’ + and from the second equation: = + = ( + ) = โˆ’ + = โˆ’ Then the desired functional relation is: = โˆ’ โˆ’ + 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 = ! โˆ’ For isothermal process, dT = 0, and ! = โˆ’ ! For ideal gas, V = nRT/P and ฮฑ = 1/T. Then = โˆ’ 1 = โˆ’

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      

๐‘‘๐น   =  ๐‘€  ๐‘‘๐‘ƒ   +  ๐‘  ๐‘‘๐‘‰  

Since  for  V    =    V  (T,  P)      

๐‘‘๐‘‰     =    ๐‘‰๐›ผ  ๐‘‘๐‘‡   โˆ’  ๐‘‰๐›ฝ  ๐‘‘๐‘ƒ  

we  have  

๐‘‘๐น     =    ๐‘€๐‘‘๐‘ƒ   +  ๐‘   ๐‘‰๐›ผ๐‘‘๐‘‡ โˆ’ ๐‘‰๐›ฝ๐‘‘๐‘ƒ    =    ๐‘๐‘‰๐›ผ๐‘‘๐‘‡   +   ๐‘€ โˆ’ ๐‘๐‘‰๐›ฝ ๐‘‘๐‘ƒ  

Again,  since  for  the  function  F  =  F  (T,  P),      

๐‘‘๐น     =    โˆ’(๐‘† + ๐‘ƒ๐‘‰๐›ผ)๐‘‘๐‘‡ + ๐‘ƒ๐‘‰๐›ฝ๐‘‘๐‘ƒ  

Equating  the  coefficients  of  dT  and  dP  from  the  above  two  equations,  we  get  

๐‘๐‘‰๐›ผ     =    โˆ’(๐‘† + ๐‘ƒ๐‘‰๐›ผ)  

๐‘€ โˆ’ ๐‘๐‘‰๐›ฝ  =    ๐‘ƒ๐‘‰๐›ฝ  

From  the  first  equation:        

๐‘     =  โˆ’๐‘† + ๐‘ƒ๐‘‰๐›ผ๐‘‰๐›ผ

 =  โˆ’๐‘†๐‘‰๐›ผ

+ ๐‘ƒ    

and  from  the  second  equation:      

๐‘€     =    ๐‘ƒ๐‘‰๐›ฝ   +  ๐‘๐‘‰๐›ฝ     =    ๐‘‰๐›ฝ  (๐‘ƒ   +  ๐‘)  

๐‘€     =    ๐‘‰๐›ฝ   ๐‘ƒ   โˆ’  ๐‘† + ๐‘ƒ๐‘‰๐›ผ๐‘‰๐›ผ

 =  โˆ’๐‘†๐›ฝ๐‘Ž

 

Then  the  desired  functional  relation  is:  

๐‘‘๐น     =    โˆ’๐‘†๐›ฝ๐‘Ž

๐‘‘๐‘ƒ   โˆ’๐‘†๐‘‰๐›ผ

+ ๐‘ƒ ๐‘‘๐‘‰  

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  

๐‘‘๐‘†     =    ๐ถ!๐‘‡๐‘‘๐‘‡   โˆ’  ๐‘‰๐›ผ  ๐‘‘๐‘ƒ  

For  isothermal  process,  dT  =  0,  and  

๐‘‘๐‘†!    =    โˆ’  ๐‘‰๐›ผ  ๐‘‘๐‘ƒ!  

For  ideal  gas,  V  =  nRT/P  and  ฮฑ  =  1/T.    Then  

๐‘‘๐‘†     =    โˆ’  ๐‘›๐‘…๐‘‡๐‘ƒ

1๐‘‡๐‘‘๐‘ƒ     =    โˆ’   ๐‘›๐‘…

๐‘‘๐‘ƒ๐‘ƒ

 

Integrating  between  the  limits  of  (S1,  10  atm)  and  (S2,  5  atm),  we  get  

โˆ†๐‘†     =    โˆ’(1  ๐‘š๐‘œ๐‘™)(8.314  ๐ฝ/๐‘š๐‘œ๐‘™  ๐พ)   ln5  ๐‘Ž๐‘ก๐‘š10  ๐‘Ž๐‘ก๐‘š

 =  5.76  ๐ฝ/๐‘š๐‘œ๐‘™  ๐พ  

 

(b)    For  adiabatic  process,  ฮดQ  =  TdS  =  0.  Thus  

โˆ†๐‘†     =    0  

 

(c)    Here  the  function  is:    S  =  S  (V,  P).  The  differential  equation  of  this  function  is  

๐‘‘๐‘†     =    ๐ถ!๐‘‡๐‘‰๐›ผ

 ๐‘‘๐‘‰   +  ๐›ฝ๐ถ!๐‘‡๐›ผ

 ๐‘‘๐‘ƒ  

At  constant  volume,  dV  =  0,  and  

๐‘‘๐‘†!    =    ๐›ฝ๐ถ!๐‘‡๐›ผ

 ๐‘‘๐‘ƒ!  

For  ideal  monatomic  gas,  ฮฒ  =  1/P,  CV  =  3R/2,  and  ฮฑ  =  1/T.    Then  

๐‘‘๐‘†     =    3๐‘…2

๐‘‘๐‘ƒ๐‘ƒ

 

Integrating  between  the  limits  of  (S1,  10  atm)  and  (S2,  5  atm),  we  get  

โˆ†๐‘†     =    3(8.314  ๐ฝ/๐‘š๐‘œ๐‘™  ๐พ)

2ln

5  ๐‘Ž๐‘ก๐‘š10  ๐‘Ž๐‘ก๐‘š

 =  โˆ’8.64  ๐ฝ/๐‘š๐‘œ๐‘™  ๐พ  

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 ฮ”U and ฮ”H 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‘‰!    =    ๐‘›๐‘…๐‘‡!๐‘ƒ!

   =    (1  ๐‘š๐‘œ๐‘™)  (0.082  ๐‘™  ๐‘Ž๐‘ก๐‘š/๐‘š๐‘œ๐‘™  ๐พ)  (273  ๐พ)

1  ๐‘Ž๐‘ก๐‘š    =    22.34  ๐‘™  

At  constant  pressure,  the  work  done  on  the  system  

๐‘Š!    =    โˆ’๐‘ƒ(๐‘‰! โˆ’ ๐‘‰!)  

โˆ’(832  ๐ฝ)  0.082  ๐‘™  ๐‘Ž๐‘ก๐‘š8.314  ๐ฝ

   =    โˆ’(1  ๐‘Ž๐‘ก๐‘š)  (๐‘‰! โˆ’ 22.34  ๐‘™)  

๐‘‰!    =    30.55  ๐‘™  

Then  the  final  temperature  

๐‘‡!    =    ๐‘ƒ!  ๐‘‰!๐‘›๐‘…

   =    (1  ๐‘Ž๐‘ก๐‘š)  (30.55  ๐‘™)

(1  ๐‘š๐‘œ๐‘™)  (0.082  ๐‘™  ๐‘Ž๐‘ก๐‘š/๐‘š๐‘œ๐‘™  ๐พ)      =    372.51  ๐พ  

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  

โˆ†๐‘ˆ     =  ๐‘„   +  ๐‘Š     =    3000   โˆ’  832  J     =    2168  J  

 

Since  for  the  ideal  gas,  dU  =  CV  dT,  then  

๐ถ! ๐‘‡! โˆ’ ๐‘‡!    =    2168  J  

๐ถ!   372.51 โˆ’ 273  ๐พ    =    2168  J  

๐ถ!  =  21.79  J/K  

And  heat  capacity  at  constant  pressure  

๐ถ!    =    ๐ถ!  +  ๐‘…     =    21.79   + 8.314     =    30.104  J/K  

Then,  the  change  in  enthalpy  of  the  system  is  

โˆ†๐ป     =    ๐ถ! ๐‘‡! โˆ’ ๐‘‡!    =    30.104   372.51 โˆ’ 273  ๐พ    =    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 0.493x10โ€“3 Kโ€“1. These values can be taken as being independent of pressure in the range 1โ€“1000 atm.

For  the  function  H  =  H  (T,  P)  

๐‘‘๐ป     =    ๐ถ!  ๐‘‘๐‘‡   +  ๐‘‰  (1 โˆ’ ๐‘‡๐›ผ)  ๐‘‘๐‘ƒ  

At  constant  temperature  

๐‘‘๐ป!    =    ๐‘‰  (1 โˆ’ ๐‘‡๐›ผ)  ๐‘‘๐‘ƒ!  

Considering  V  of  solid  does  not  vary  too  much  with  P,  integrating  

โˆ†๐ป!    =    ๐‘‰  (1 โˆ’ ๐‘‡๐›ผ)  (๐‘ƒ! โˆ’ ๐‘ƒ!)  

=     7.09  ๐‘๐‘ 1 โˆ’  298  ๐‘ฅ  0.493๐‘ฅ10!! 1000 โˆ’ 1  ๐‘Ž๐‘ก๐‘š  

=    โˆ’6042.33  ๐‘๐‘  ๐‘Ž๐‘ก๐‘š  ๐‘ฅ  8.314  ๐ฝ

82.06  ๐‘๐‘  ๐‘Ž๐‘ก๐‘š  =  โˆ’612.19  ๐ฝ  

The  same  amount  of  enthalpy  change  is  produced  when  the  system  undergoes  a  constant  pressure  process.  Thus  

โˆ†๐ป!  =   ๐ถ!  ๐‘‘๐‘‡!

!!    

โˆ’612.19  ๐ฝ     =     22.6   +  5.6๐‘ฅ10!!  ๐‘‡  ๐‘‘๐‘‡!

!"#  

โˆ’612.19     =    22.6  ๐‘‡   +  2.8๐‘ฅ10!!  ๐‘‡!  โˆ’  22.6  (298)  โˆ’  2.8๐‘ฅ10!!  (298)  

2.8๐‘ฅ10!!  ๐‘‡!  +    22.6  ๐‘‡   +  6371.26     =    0  

๐‘‡     =    โˆ’22.6   ยฑ   22.6 ! โˆ’ 4  (2.8๐‘ฅ10!!)(6371.26)  

2  (2.8๐‘ฅ10!!)    =    292.51  ๐พ  

Hence  the  temperature  is  292.51  K.