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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      

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.