suppose 10.0g of dry ice (solid CO2) was placed in an empty 400-mL steel cylinder.What pressure would develop if all the solid sublimed at a temperature of 35.0*C?
2006-10-13 11:04:16 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
First you need to convert grams of Carbon Dioxide to moles of Carbon Dioxide using the molar mass of CO2.
CO2 has a molar mass of 44.01 g/mol.
10 grams of CO2 equals 10 g / 44.01 g/mol = .227 moles of CO2.
Now use the ideal gas law (we assume CO2 acts as an ideal gas),
PV = nRT
You know the volume (V) of the container, you also know the number of moles (n), gas rate constant (R), and temperature (T) of the gas at equilibrium.
All you need to do is solve for the pressure (P).
Remember, the gas rate constant will depend upon the units you use for the other variables and will effect what units you final pressure will be in. Also, the temperature (T) is in Kelvin, not Celsius. Convert degrees C to K, add 273.15.
Based on my own calculations, I find the answer to be 14.36 atmospheres of pressure.
V = .400 Liters
n = .227 moles
R = .0821 Liter atm /mol K
T = 308.15 K
2006-10-13 11:12:20 · answer #1 · answered by mrjeffy321 7 · 2⤊ 0⤋
They pressure wouldn't change or anything. It would stay constant. The pressure would have been the kPA that is at water level. Carbon dioxide cannot exist as a liqiud at that pressure. Therefore, it would sublimate.
2006-10-13 18:11:46 · answer #2 · answered by symperl 2 · 0⤊ 1⤋
PV=nRT P= (M/Mw)*R*(273+35)/0.4
2006-10-13 19:18:26 · answer #3 · answered by maherrashdan 2 · 0⤊ 0⤋
its too tough.i will give it to u tomorrow
2006-10-14 13:10:52 · answer #4 · answered by MANSI R 2 · 0⤊ 0⤋