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CO2 gas in a temp-controlled, pressure-controlled container |
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| Jan28-11, 12:37 AM | #1 |
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CO2 gas in a temp-controlled, pressure-controlled container
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 |
| Jan28-11, 12:46 AM | #2 |
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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? |
| Jan28-11, 12:51 AM | #3 |
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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. |
| Jan28-11, 01:07 AM | #4 |
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CO2 gas in a temp-controlled, pressure-controlled container
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. |
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