Solving Adiabatic Process Problem: Find Final Volume & Pressure

In summary, the gas expanded adiabatically until the pressure was halved, find the final volume and final pressure. Using the ideal gas equation, PV=nRT, the final volume is 1.37 x 10^-5 m^3.
  • #1
Spectre5
182
0
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?
 
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  • #2
Spectre5 said:
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.

[tex] P_{initial}/P_{final}=(V_{final}/V_{initial})^{1.4}=2[/tex]

[tex]V_{final}/V_{initial}=2^{1/1.4}=1.641[/tex]

[tex] V_{final}=1.970 \cdot10^{-3}\mbox{ } m^3[/tex]

ehild
 
  • #3
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 :)
 
Last edited:
  • #4
Spectre5 said:
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
 

1. What is an adiabatic process?

An adiabatic process is a thermodynamic process in which no heat is transferred between the system and its surroundings. This means that the change in internal energy of the system is equal to the work done on the system.

2. How do you solve an adiabatic process problem?

To solve an adiabatic process problem, you first need to determine the initial volume and pressure of the system. Then, using the adiabatic process equation: P1V1^γ = P2V2^γ, where γ is the ratio of specific heats, you can calculate the final volume and pressure of the system.

3. What is the ratio of specific heats?

The ratio of specific heats, also known as the adiabatic index, is a thermodynamic property that describes the relationship between the specific heats at constant pressure and constant volume. It is denoted by the symbol γ and is typically between 1 and 2 for most gases.

4. How does temperature change in an adiabatic process?

In an adiabatic process, the change in temperature depends on the work done on or by the system. If work is done on the system, the temperature will increase. If work is done by the system, the temperature will decrease. However, the change in temperature will not be equal to the change in volume, as it would be in an isothermal process.

5. What are some real-world applications of adiabatic processes?

Adiabatic processes have many practical applications, including in the compression and expansion of gases in engines, the operation of refrigerators and heat pumps, and in the formation of weather patterns such as thunderstorms. They are also commonly used in industrial processes, such as in the production of compressed air and in gas turbines.

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