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Work, entropy, and the ideal gas law

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A balloon filled with 40 moles of He at 3 atm pressure and at room temperature (300 K) expands isothermally from a volume Vi to a volume Vf till its pressure falls to 1 atm.

(a) What are Vi and Vf?

Here's what I did for this one:

PV = nRT

Vi = nRT/P = (40)(8.315)(300K)/[3*(1.013 x 105 N/m2) = 0.33 m3

Vf = nRT/P = (40)(8.315)(300K)/(1.013 x 105 N/m2) = 0.98 m3

Does this look correct to you?

(b) What is the work done during the expansion?

Here's what I did for this one:

W = nRT ln Vi/Vf = (40)(8.315)(300) ln 0.33/0.98 = 108606.5 J

Does this look correct?

(c) What is ΔU?

I know ΔU = W - Q. If part (b) is correct, I know that W = 108606.5 J, but what is Q? How do I calculate Q?

(d) What is Q (heat exchanged) during the process?

This is just a reiteration of my own previous question, as without Q, I don't know how to calculate ΔU.

(e) Calculate ΔS directly from the Sackur-Tetrode equation.

I am uncertain as to what variables go with what, and still cannot do this since I don't know Q.

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https://brainmass.com/physics/internal-energy/work-entropy-ideal-gas-law-104385

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You did a) correctly except for the units for the gas constant. You should include a Joule/Kelvin in the factor 8.315. So you can write 40 Mole times ( 8.315 J/(K Mole)) = 40 *8.315 J/K

In b) what you should calculate is the work done by the gas. You wrote down the formula for work done on the gas. And then you made an error with the sign because the logarithm you wrote down is negative.

The work done by the gas is the integral of P dV from the initial state to the final state and is given by:

W = nRT Log[Vf/Vi]

When you work out ...

Solution Summary

A detailed solution is given. The expert calculates entropy from the Sackur-Tetrode equations. The processes during the heat exchange are determined.

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