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asim14990 asim14990
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11 years ago
A 1.30g sample of dry ice is added to a 765mL flask containing nitrogen gas at a temperature of 25.0C  and a pressure of 735 mmHg. The dry ice is allowed to sublime (convert from solid to gas) and the mixture is allowed to return to 25.0 C.
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11 years ago
Use the ideal gas equation:
PV = nRT

Find the pressure of the dry ice gas that exists when
V = 765mL = 0.765L
T = 25 + 273 = 298K
R = 0.08206 Latm/molK

Also remember that dry ice is the solid form of carbon dioxide, CO2. Find the number of moles present by multiplying the weight by the molar mass for CO2:
n = 1.30g x (1 mol/44.010g) = 0.0295 mol

Now use the equation and solve for P:
P = nRT/V = (0.0295mol)(0.08206Latm/molK)(298K) / (0.765L)
P = 0.0943 atm

This is the pressure that the dry ice will exert (in atmospheres). The total pressure in the flask will be equal to the sum of all pressures (the pressure of nitrogen plus the pressure of CO2 gas), but we are given the pressure of nitrogen gas as 735 mmHg, and we found the pressure of the CO2 in atmospheres. You must convert either one to the same unit to add them together. Let's convert the pressure we found of the CO2 gas into mmHg units.
0.0943 atm x (760 mmHg/atm) = 717 mmHg

Now we can add the two pressures together:
Ptotal = 717 mmHg + 735 mmHg = 1452 mmHg = 1.91 atm
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