Calculating Pressure & Temp for NH3 Quasistatic Adiabatic Expansion

[forty]

(a)
For a quasistatic adiabatic process involving an ideal gas the temperature and volume are related by: TV^(a-1) = const . By substituting the ideal gas equation into this expression, derive a similar relationship between pressure and volume for an ideal gas in an quasistatic adiabatic process.

Using PV = nRT and rearranging to get PV/nR=T

Sub into TV^(a-1) = const --------> (PV/nR)V^(a-1) = const

(PV^a)/nR = const

(b)
In one of the lecture demonstrations it was shown that the temperature of CO2 decreases rapidly as it expands upon release from a fire extinguisher. We could model this process as a quasiadiabatic expansion of an ideal gas to calculate a “ball-park” value for the temperature change of the CO2. If we wanted to use a similar process to cool ammonia, NH3, to its boiling point at 1 atmosphere pressure (a value that you may need to look up) find the initial pressure and temperature that would be required if the volume occupied by the gas increases by a factor of 10 in the expansion process, i.e., final values: pressure p0 = 1 atm, volume = V0 , temperature = boiling point of NH3, initial values: pressure pi = ?, volume Vi = V0/10, temperature Ti = ?. You may need the following information for NH3: Cp/Cv = 1.31.

Ammonia boiling point = 240k at 1 atm.

Vi = Vf/10
Pi =
Ti =

Pf = 1atm
Tf = 240k
Vf = Vf

Using the ideal gas law for each then dividing i get Pi/10atm = Ti/240k

But after this i don't know where to go... i tried using the results from (a) to get Ti or Pi in terms of the other but had no real success.

Any help or ideas would be greatly appreciated.

(Cp/Cv = 1.31 <---- what does this actually mean?)

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EDIT:

I realized that Cp/Cv = 1.31 = a

then solved using the result from part a and got Pi = 20.42 atm and Ti = 490.02k

 

[Jhero]

To find the initial pressure and temperature for ammonia, we can use the relationship derived in part a: (PV^(a))/nR = const.

Substituting the values given for ammonia, we get:

(Pi*Vi^(a))/nR = (Pf*Vf^(a))/nR

Since we know that Vf = V0 and Vi = V0/10, we can simplify the equation to:

(Pi*(V0/10)^(a))/nR = (1*V0^(a))/nR

Solving for Pi, we get:

Pi = 10^(1/a) = 10^(1/1.31) = 6.3 atm

To find Ti, we can use the ideal gas law: PV = nRT

Solving for T, we get:

T = (PV)/(nR)

Substituting the values given for ammonia, we get:

T = (6.3 atm * V0/10)/(1 mol * 0.0821 L atm/mol K) = 0.766 V0 K

Since we know that the final temperature is 240 K, we can set up an equation and solve for V0:

0.766 V0 K = 240 K

V0 = 313.2 L

Therefore, the initial pressure and temperature required for the ammonia to reach its boiling point at 1 atm pressure after expanding by a factor of 10 are 6.3 atm and 313.2 L, respectively.