How can I calculate final pressures in a reversible argon expansion?

In summary, the conversation revolves around solving a problem involving one mole of argon at 25 degrees C and 1 atm pressure expanding to a volume of 50 L, both isothermally and adiabatically. The final pressure in each case is calculated using ideal gas behavior and the adiabatic condition. The final answers are given as 0.489 atm for the isothermal expansion and 0.303 atm for the adiabatic expansion. The conversation also includes a discussion of using Boyle's law and the ideal gas law to solve the problem.
  • #1
marissy
2
0
isothermal and adiabatic please help!

Homework Statement


Hello,

I have been stuck on this problem for about an hour so any hep would be greatly appreciated!

One mole of argon at 25 degrees C and 1 atm pressure is allowed to expand reversibly t a volume of 50 L (a) isothermally (b) adiabatically. Calculate the final pressure in each case assuming ideal gas behavior.
The answers have been given and they are: (a) 0.489 atm and (b) 0.303 atm

I can't get either a or b for some reason I don't know what I am doing wrong! Please help me :(


Homework Equations


Cv = 3/2 R Cp= 5/2 R
Boyle's law


The Attempt at a Solution



my answers don't even come close...?
 
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  • #2


Show some work please. It's impossible to show you where you went wrong if you don't show us what you have done.
 
  • #3


nevermind i got it all -- I just didnt have my P and V in the right units... anyway i found V1 using the ideal gas law, and the used boyle's law to find P2. Seems so simple when you have the correct units :)

for part b i did PV(5/3)=nRT to find V1 and then used boyle's law to find P2. I got 0.296 for the answer tough. maybe my sig. figs. are off? i can't find out why.
 
  • #4


marissy said:
for part b i did PV(5/3)=nRT to find V1 and then used boyle's law to find P2. I got 0.296 for the answer tough. maybe my sig. figs. are off? i can't find out why.
You will have to use the adiabatic condition:

[tex]PV^\gamma = \text{constant} \ne nRT[/tex]

for part b). [itex]PV^1 = nRT[/itex]

AM
 
  • #5


Yeah, as AM has said, use the adiabatic condition, along with energy conservation to figure it out.
 

FAQ: How can I calculate final pressures in a reversible argon expansion?

1. What is the difference between isothermal and adiabatic processes?

The main difference between isothermal and adiabatic processes is that isothermal processes occur at a constant temperature, while adiabatic processes do not exchange heat with the surroundings. In other words, isothermal processes maintain a constant internal energy, while adiabatic processes may experience changes in internal energy due to work done on or by the system.

2. How do you calculate the work done in an isothermal or adiabatic process?

The work done in an isothermal process can be calculated using the formula W = nRT ln(V2/V1), where n is the number of moles of the gas, T is the temperature, and V2 and V1 are the final and initial volumes, respectively. For adiabatic processes, the work done can be calculated using the formula W = (P2V2 - P1V1)/(1-gamma), where P is the pressure, V is the volume, and gamma is the adiabatic index.

3. What is the ideal gas law and how does it relate to isothermal and adiabatic processes?

The ideal gas law is a formula that describes the relationships between pressure, volume, temperature, and number of moles of a gas. It is expressed as PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. In isothermal processes, the ideal gas law holds true as the temperature remains constant. In adiabatic processes, the ideal gas law is modified to account for changes in temperature.

4. What are some real-life examples of isothermal and adiabatic processes?

An example of an isothermal process is the expansion of a balloon in a room with constant temperature. The air inside the balloon remains at a constant temperature as it expands, allowing the ideal gas law to apply. An example of an adiabatic process is the compression of a gas in a bicycle pump. As the gas is compressed, it experiences an increase in pressure and temperature, but since it is insulated, it does not exchange heat with the surroundings.

5. How do isothermal and adiabatic processes affect the efficiency of a system?

In general, isothermal processes are more efficient than adiabatic processes. This is because in an isothermal process, the energy is conserved and there is no change in internal energy, while in an adiabatic process, some of the energy is lost as heat. However, adiabatic processes can also be more efficient in certain cases, such as in the compression of gases in engines, where the increase in temperature can lead to a more efficient combustion process.

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