CO2 gas in a temp-controlled, pressure-controlled container

Click For Summary

Homework Help Overview

The problem involves calculating the gas pressure and final temperature of carbon dioxide in a sealed container after it undergoes phase changes and compressions. The subject area includes thermodynamics and gas laws.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to relate the latent heat of vaporization to the pressure of CO2 but expresses uncertainty about the next steps and relevant equations. Other participants calculate pressure using the ideal gas law but struggle with unit conversions and signs in their calculations.

Discussion Status

Participants are actively exploring different approaches to calculate the pressure and temperature. Some have provided calculations, but there is a lack of consensus on the correct method and results. The discussion is ongoing with attempts to clarify the use of units and constants.

Contextual Notes

There is a noted confusion regarding the application of the ideal gas law and the correct temperature conversion, as well as the implications of the phase change from solid to gas.

mcnivvitz
Messages
7
Reaction score
0
The problem:
20 g of dry ice (solid CO2) is placed in a 2.0*10^4 cm^3 container, then all the air is quickly pumped out and the container sealed. The container is warmed to 0 deg C, a temperature at which CO2 is a gas.

a) What is the gas pressure? Give your answer in atm. The gas then undergoes an isothermal compression until the pressure is 3.4 atm, immediately followed by an isobaric compression until the volume is 2000 cm^3.

b) What is the final temperature of the gas?

My thoughts:
Something to do with latent heat of vaporization. For carbon dioxide, this constant is 574 kJ/kg. 20 g = .2kg of CO2. No idea how this translates into pressure. Where do I go from here? Can anyone give me some relavent equations?

Thanks
 
Physics news on Phys.org
P = nRT / V
Molar mass of CO is 28.010 g mol^-1.
Therefore number of moles in 20 g is 20 g/28.010 g mol^-1 = 0.71 mol.

P = ((.71)(8.314472e15)(-273.15)) / (2.0*10^4 cm^3)

P = -8.062398*10^19 m^3

Is this correct? What do I do now?
 
Err, I put the wrong units and gas constant. Here:

((.71 mol)(8.3145 J/mol K)(-273.15 degrees K)) / (2.0*10^4 cm^3) = P = -80624251.5 kg K / s2

That's not right either. Sigh.
 
Did the calculation wrong. Now I've got

((.71 mol)(8.3145 J/(mol*kelvin))(-273.15 degrees kelvin)) / (2.0*10^4 cm^3) = -0.795699496 atm

Not the right answer for pressure. But at least the proper units.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
11
Views
5K
  • · Replies 12 ·
Replies
12
Views
5K