A 1.50 rubber balloon balloon is filled with carbon dioxide gas at a temperature of 0.00 degrees celsius and a pressure at 1.00 atm. The density of carbon dioxide under these conditions are 1.98 g/L.
At 50.0 degrees celsius, the balloon has a volume of 1.78 L. Calculate the carbon dioxide density at this temperature.
Please help!!! And show clears steps, thanks!!
4 answers
I'm sorry I meant to put a 1.50 L of rubber ballon!
Step 1:
Considering the gas equation for the first condition:
PV = nRT
n = PV/RT
= 1*1.5/0.0821*273
= 1.5/22.41
= 0.066 mol
So we now know that the number of moles of CO2, which will remain fixed, is 0.066 mol.
Step 2:
In the new case -
Total mass of CO2 = Molar mass * number of moles
= 44 * 0.066
= 2.904
Density = Mass/Volume
= 2.904g/1.78L
= 1.63g/L
Considering the gas equation for the first condition:
PV = nRT
n = PV/RT
= 1*1.5/0.0821*273
= 1.5/22.41
= 0.066 mol
So we now know that the number of moles of CO2, which will remain fixed, is 0.066 mol.
Step 2:
In the new case -
Total mass of CO2 = Molar mass * number of moles
= 44 * 0.066
= 2.904
Density = Mass/Volume
= 2.904g/1.78L
= 1.63g/L
I agree with the way the problem is worked above; however, I would round the 0.066 mols to 0.067 which makes the new density at 323 K to be 1.66 g/L which to to two significant figures is essentially the same.
Here is much shorter way to work the problem. Since density = g/L and grams is constant, then density varies as T. If T goes up we know volume goers up and a larger volume means a smaller density. So 1.98 g/L x (273/323) = 1.67 g/L.
Here is much shorter way to work the problem. Since density = g/L and grams is constant, then density varies as T. If T goes up we know volume goers up and a larger volume means a smaller density. So 1.98 g/L x (273/323) = 1.67 g/L.
Thank you very much Arora and DrBob!! You guys were very helpful :)
1 answer