Volume of Balloon at 23K and 299K

[jagged06]

Homework Statement

Imagine that 9.9 g of liquid helium, initially at 4.20 K, evaporate into an empty balloon that is kept at 1.00-atm pressure. What is the volume of the ballonat the following?

(a) 23.0 K(b) 299 K

Homework Equations

PV=nRT

The Attempt at a Solution

I converted th 9.9g of Helium to number of moles. n=2.473mol
and used that to find the volume at T=23 and T=299, but it isn't right.

steps
1) n=9.9g=(9.9g/(4.003g/mol))=2.47314 mol
R=8.314472 <-- gas constant
P= 1 as stated in the problem

V@ T=23 : V=(nRT)/P -> (2.47314*8.314472*23)/1 --> V=472.9466 L

V@ T=299 : V=(nRT)/P -> (2.47314*8.314472*299)/1 --> V=6148.30591237 L
I'm guessing I need to do something with T=4.2 first, but I'm not sure what.

 

[kuruman]

Please show your work, i.e. exactly what you did, and then perhaps we can determine where you went wrong.

 

[jagged06]

edited to show work

 

[kuruman]

Your problem is a mixed bag of units. When you write pV = nRT
p is measured in Pa (or N/m²) not atmospheres. How many Pa is one atmosphere? Your textbook should have the number. When you put in the correct pressure, the answer should come out in m³. If you want liters you need to make another conversion.

 

[rdl]

I would like to point out that the ideal gas law, PV=nRT, is only applicable to gases at low pressures and high temperatures. At the temperature and pressure given in the problem (4.20 K and 1.00 atm), helium is not in a gaseous state but rather in a liquid state. Therefore, the ideal gas law cannot be used to calculate the volume of the balloon at 23 K and 299 K.

To solve this problem, we need to use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature and enthalpy of vaporization. Since the problem states that the helium is evaporating into an empty balloon at 1.00 atm, we can assume that the pressure inside the balloon is also 1.00 atm. Therefore, the vapor pressure of helium at both 23 K and 299 K will be equal to 1.00 atm.

Using the Clausius-Clapeyron equation, we can solve for the volume of the balloon at both temperatures. At 23 K, the vapor pressure of helium is 1.00 atm and its enthalpy of vaporization is 2.08 kJ/mol. Plugging these values into the equation, we get:

ln(1.00/1.00) = (-2.08 kJ/mol)/(8.314 J/mol*K * 23 K) + C

Solving for C, we get C = 0. Therefore, at 23 K, the volume of the balloon is equal to the volume of the evaporated helium, which is 2.473 mol.

At 299 K, the vapor pressure of helium is still 1.00 atm, but its enthalpy of vaporization is now 21.2 kJ/mol. Plugging these values into the equation, we get:

ln(1.00/1.00) = (-21.2 kJ/mol)/(8.314 J/mol*K * 299 K) + C

Solving for C, we get C = 0. Therefore, at 299 K, the volume of the balloon is equal to the volume of the evaporated helium, which is 2.473 mol.

In conclusion, the volume of the balloon at both 23 K and 299 K is equal to 2.473 mol, which is equivalent to 472.9466 L