Solving Adiabatic Process Problem: Find Final Volume & Pressure

[Spectre5]

I have this problem and I do not have the answer, but I get an answer that I feel is probably wrong, so can someone please check my work and point out where I went wrong??

Here is the problem:
moles = 0.10 of O_2
T(initial) = 150 C = 423 K
P(initial) = 3.0 atm = 303.9 KPa

The gas expands adiabatically until the pressure is halved, find the final volume and final pressure

Since the pressure is halved, we know that
P(final) = 1.5 atm = 151.95 KPa

I need the initial volume, so I used the ideal gas equation, PV = nRT
using the initial conditions with P in pascals, n in mols, T in kelvin, and R as 8.31 J/mol*K

So I get a V(initial) = 1.156 x 10^(-3) m^3

Then I need the final volume, and since this is adiabatic,
Pi(Vi)^(gamma)=Pf(Vf)^(gamma)

Since O_2 is diatomic and we assume ideal conditions, gamma = 1.4

So using the above equation, I find
V(final) = 1.37 x 10^(-5) m^3 = answer to part a
I don't know if this is right or wrong

Then for part b, I used the idea gas equaion again, PV=nRT
Using the final volume, final pressure, same n and same R, I get
T = 2.50 K

Obviously this is extremely COLD! I don't think it makes sense that the temperature would drop from 423 K to 2.5 K...where did I go wrong?

 

[ehild]

So I get a V(initial) = 1.156 x 10^(-3) m^3

Then I need the final volume, and since this is adiabatic,
Pi(Vi)^(gamma)=Pf(Vf)^(gamma)

Since O_2 is diatomic and we assume ideal conditions, gamma = 1.4

So using the above equation, I find
V(final) = 1.37 x 10^(-5) m^3 = answer to part a
I don't know if this is right or wrong

The final volume can not be right, as it is much lower than the initial volume, and the gas has expanded.

$$P_{initial}/P_{final}=(V_{final}/V_{initial})^{1.4}=2$$

$$V_{final}/V_{initial}=2^{1/1.4}=1.641$$

$$V_{final}=1.970 \cdot10^{-3}\mbox{ } m^3$$

ehild

 

[Spectre5]

Yes, I just realized that at the same time you posted...

I used the wrong initial volume (actually I just used 1.156 instead of 1.156 x 10^-3

:/

thanks

btw..it is 1.90 x 10^-3 I think, not 1.970...probably just a typo though :)

 

[ehild]

btw..it is 1.90 x 10^-3 I think, not 1.970...probably just a typo though :)

Well, yes, it was 1.897 and I left out the "8" :)

ehild