How Does Helium's Entropy Change After Slow Isothermal Escape?

[mewmew]

Here is the problem:
"A careless experimenter left the valve of a tank of helium slightly open over the weekend. The gas, originally at 200 atm, slowly escaped isothermally at 20 degrees C. What change in entropy per kg of gas occured? " I am given, K, N, and Mass of helium per mole.

I am stuck because all I know is:

Delta S = N k Ln[Vf/Vi]
and
P=NkT/V

I can't seem to figure out how to make these work though given the problem at hand. How do I find out what N-final and V-final are? That is assuming I even need to find those out to solve this problem. Any help is greatly appreciated!

 

[Andrew Mason]

Here is the problem:
"A careless experimenter left the valve of a tank of helium slightly open over the weekend. The gas, originally at 200 atm, slowly escaped isothermally at 20 degrees C. What change in entropy per kg of gas occured? " I am given, K, N, and Mass of helium per mole.

I am stuck because all I know is:

Delta S = N k Ln[Vf/Vi]
and
P=NkT/V

I can't seem to figure out how to make these work though given the problem at hand. How do I find out what N-final and V-final are? That is assuming I even need to find those out to solve this problem. Any help is greatly appreciated!

Use dS = dQ/T

$$\Delta S = \int ds = \int dQ/T = \Delta Q/T$$ where the heat flow occurs at the same temperature.

You just have to find the heat flow in expanding 200 times (ie pressure going from 200 atm to 1). Use the first law:

$$dQ = dU + PdV = nC_vdT + PdV$$

since T is constant, dU = 0 so dQ = PdV

So the heat flow out of the gas is equal to the work done by the gas. What is the work done by the gas?

AM