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Volume increasing and work?The balloon contains 0.30 mol of helium starts at a pressure...
Volume increasing and work?
The balloon contains 0.30 mol of helium starts at a pressure of 125kPa and rises to an altitude where the pressure is 70.0 kPa, maintaining a constant 300 K temperature.
By what factor does its volume increase? v2/v1=
How much work does the gas in the balloon do? W= J
1 Answer | add yours
To find the value of pressure-volume work, we need to look at the formula
w = - P (deltaV)
Where P is the external pressure and deltaV is the change in volume. To find the change in volume, we need to use the ideal gas law to find the initial pressure and the final pressure. First we need to convert both pressures to atm
125 kPa (1 atm / 101.3 kPa) = 1.23 atm
70.0 kPa (1 atm / 101.3 kPa) = 0.691 atm
Now we need to find the volume at each pressure using the ideal gas las.
PV = nRT
(1.23 atm)V = 0.30 (0.08206 L atm /molK)(300)
V = 6.00 L
and at the lower pressure
(0.691 atm)V = 0.30 (0.08206 L atm/mol K) (300K)
V = 10.7 L
Now, we can find deltaV is 10.7-6.00 = 4.7 L. So the balloon expands 4.7 L.
To find the ratio of the volumes, we can simply take V2/V1 = 10.7/6.00 = 1.78
To find the work, we use the work equation
w = -P(deltaV)
w = -0.691 atm(4.7 L)
w = - 3.24 L atm
Convert L atm to J by multiplying by 101.3 J
-3.24(101.3) = -328 J
Posted by mlsiasebs on April 5, 2012 at 10:44 AM (Answer #1)
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