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# What is the partial pressure of CO2? What is the total pressure in the container?

Given

Dry ice weighs 2g in a sealed 1L container filled with air (25 degrees celsius) @ STP. The ice sublimes and the temperature returns to 25 degrees.

My answer

V=1L

T=298K

n=22.045

R=8.31

PV=nRT

22.045 X 8.31 X 298

P=54592

Is this correct?

P total = P(air) + P (Co2)

+ 101.3 Kpa + 111.4 Kpa

=212.7 Kpa or 2.1 atm correct?

### 2 Answers

- Dr WLv 71 decade agoFavorite Answer
total pressure = partial pressure CO2 + partial pressure air

partial pressure air = 1 atm...(STP)

**** actually, STP is OC and 1 atm.. not 25C and 1 atm... 25 C and 1 atm is what the US EPA uses for standard conditions.. see the source...regardless. 25C and 1 atm it is...*****

so let's find PCO2...

PV = nRT

P = nRT/V

n = 2 g CO2 x (1 mole CO2 / 44.0 g) = 0.004535 moles

R = 0.0821 Latm/moleK

T = 25C = 298.15K...

V = 1 L

PCO2 = (0.004535 moles) x (0.0821 Latm/moleK) x (298K) / (1 L) = 0.1 atm... 1 sig fig since 2g has 1 sig fig..

total pressure = pressure air + pressure CO2 = 1.1 atm.

********

your calculation of moles was incorrect.

my other suggestion is that you follow through on the units..

- pisgahchemistLv 71 decade ago
No, 54592 atm is absurdly high.

You don't have 22.045 moles, you have 0.04545 mol of CO2

R can be 8.314 LkPa/molK if you want pressure in kPa, or you can use 0.0821 Latm/molK to get the pressure in atm.

PV = nRT

PV = mRT / M where m = mass, M = molar mass

P = mRT / VM

P = 2.00g x 0.0821 Latm/molK x 298K / 1.00L / 44.0g/mol = 1.11 atm