Entropy, Internal Energy, etc

Question

Consider a perfect gas contained in a cylinder and separated by a frictionless adiabatic piston into two sections A and B. All changes in B are isothermal. There is 2 mol of the gas in each section. Initially temperatures in both sections are 300K, and the volume of both sections are 2dm^3. Energy is supplied as heat to section A and the piston moves to the right reversibly until the final volume of section B is 1dm^3. Calculate (a) the heat q(A) added to section A, (b) the work w(A) done by section A, (c) change in internal energy for both sections, (d) change in entropy for both section, and for the total system and surroundings. If numerical values cannot be obtained, indicate whether the values should be positive, negative, or are indeterminate from the info given. (Assume Cv,m=20J/Kmol)

Please be descriptive, if an answer cannot be determined please explain why, and be as concise as possible please, list all steps. Thank you

Answer 1

Consider gas B:

dU = dq + dw

 

since processes for B are isothermal, then dU = 0 and

 

0 = dq + dw

dq = -dw

 

for an ideal gas we know that:

 

dw = -PdV = -(nRT/V)dV

 

integration gives us:

 

w = -nRT[ln(Vf/Vi)]

 

thus,

 

q = nRT[ln(Vf/Vi)]

 

plug in all the given values into the equation above to determine how much heat and/or work was been added/done on system B. Assuming that all the heat input in A was used for the expansion then qa (heat of A) must be:

 

qa = -qb

 

work of A can be determine using the same equation we derived above:

 

w = -nRT[ln(Vf/Vi)]

 

the internal energy of A is:

 

dU = dqa + dw

 

dU = -dqb -nRT[ln(Vf/Vi)]

 

alternatively you can determine dU using the heat capacity since

 

dU = CvdT

 

or

 

U = Cv(Tf - Ti)

 

you will have to determine the final temperature using the ideal gas law on A after it has expanded.

 

At this point I will leave the determination of the entropy for you to figure out. Just remember that dS = dq/T.