Solving Thermo Difficulties: Calculating Entropy Change

[Perses]

Hello; I'm having some difficulties with a thermo question of mine.

at low density water vapor conforms well to the ideal gas equation provided the temp is higher than about 320K, but the heat capacity is a function of temperature. The following formula gives the specific heat capacity at constant volume, as a function of T
cv = 1273.0 + 0.3441T + (2.833x10^-4)T^2 J/kgK

a) calculate the entropy change for 1kg of water vapour heated from 350K to 1000K at constant volume.

Here's what i did; S = (integral)dQ/T = (integral)NcvdT/T. Where N = (m/MM) = 1/0.018
I subbed cv into this equation then divided it by T
= (m/MM)(1273.0/T + 0.3441 + (2.833x10^-4)T)dT

Then integrated over T1 = 350, T2 = 1000 to get:

S = (1/0.018)(1273.0(ln(1000/350)) + 0.3441(1000-350) + ((2.833x10^-4)/2)(1000-350)^2)

however, my answer is not correct according to the back of the book. The answer is supposed to be 1.68kJ/K. Where did i go wrong?

b) Same as a) except at constant pressure.
here's what i tried;
S = (inegral)(Ncpdt)
where cp = cv + R

hwoever this also doesn't give me the correct answer.

Any help or suggestions would be greatly appreciated!

 

[uart]

You're on the right track, just two simple mistakes in the first part.

1. There's no need to multiply by "N = (m/MM) = 1/0.018" as the specific heat is already in J/Kg/K. (note per Kg not per mole). Alway check the units of the constants you're using and make sure they conform to the units you're using in the rest of the problem.

2. You made a very simple mistake with the "^" squared part when subtituting in the limits of integration. I'll let you have another look and find this mistake yourself. :)

 

[Perses]

Ah, yes! thank you very much. What a dumb mistake!

Anyways; i was wondering if anyone could give me a starting point with this next problem!

This is the question that the problem i'mhaving trouble with refers to. I can do this one easily:
two cylinders, each containing an ideal monatomic gas, are fitted with fritcionless pistons. The apparatus is constructed so that the two pistons and the 10kg mass they support, always move together and by equal amounts. The weight of the pistons and push rod assembly is negligible, the area of the piston A is 0.02m2, and the area of piston B is 0.01m2. Take g = 10m/s2 and external pressure to be zero.
Initially the push rod assemgly is clamped with xa = 0.01m and xb =0.3m where xa and xb are the height of the piston A and B respectively. Both cylinders the temperature is 300K and the pressure is 1x10^4Pa. When the clamp is released, the system is allowed to move freely until a new state of eq is reached. The temp of the gas is in both cylinders is 300K

Here is the problem i can't seem to start properly:
A system is set up as described above, however, when The clamp is released the motion of the piston yoke is restrained by some external force so that the expansion process takes place quasistatically and adiabatically.
Determine the equilibrium temp and pressure in each cylinder and the amount by which the 10kg mass is elavated.

I'm not quite sure were to go with this. I thought that the force on each piston was half of the force the weight was pushing down with, this is what i did in the upper question and got the right ansewr, but in this part the pressures are different.