Thermodynaics Analysis of Processes

[AriAstronomer]

Homework Statement

A cylinder, A, having Va=30L, is connected to a second cylinder B, of circular cross section. Cylinder B is fitted with a piston of radius = 5cm, which slides without friction/leaking. Initially A contains an ideal monatomic gas, in addition, Pa = 10MPa, Ta = 300K, Vb = 0 and the compression force on the spring is zero. The spring, obeying Hooke's law has a force constant of 10^5N/m. The valve, V, is opened slowly and gas leaks from A to B, compressing the spring. Equilibrium is eventaully established again at a temperature of 300K throughout all parts of the system. Neglect the volume of the connecting pipe (between A and B).

a) Calculate the distance by which the spring is compressed in the final equilibrium state.

Homework Equations

Vtotal = Vo - xA where x is the distance the piston has moved.
F=PA = -kx

The Attempt at a Solution

So the answer in the book is .6684m. But I keep getting .7854m:
Vtotal = 30L - xA = .03m^3 - xA
Thus: x = (.03m^3 - V)/A
plug into PA = -kx and solve for V, which I get Vtotal = .02383m^3.
Then plug Vtot back into original Vtot equation and get
x = .7854m.
But the answer in the back says .6684m

Any help would be appreciated.

 

[nate808]

Thank you for your question. It seems like you are on the right track in your solution, but there may be a small error in your calculations. Here is how I would approach the problem to get the correct answer:

First, let's define our variables:

V_A = volume of cylinder A = 30 L = 0.03 m^3
V_B = volume of cylinder B = 0 m^3 (initially)
r = radius of piston = 5 cm = 0.05 m
P_A = pressure of gas in cylinder A = 10 MPa = 10,000,000 Pa
T_A = temperature of gas in cylinder A = 300 K
k = force constant of spring = 10^5 N/m

Now, let's use the ideal gas law to relate the initial and final states of the gas in cylinder A:

P_A V_A / T_A = P_B V_B / T_B

Since V_B = 0 and T_B = T_A = 300 K, we can solve for P_B:

P_B = P_A V_A / V_B = 10,000,000 * 0.03 / 0 = undefined

This makes sense, since we know that as the gas leaks into cylinder B, it will compress the spring and reach equilibrium at a lower pressure than 10 MPa. So let's use the work-energy theorem to relate the change in potential energy of the spring to the work done by the gas in expanding:

W = ΔU = U_f - U_i

where W = work done by gas, ΔU = change in potential energy of spring, U_f = final potential energy of spring, and U_i = initial potential energy of spring.

We can also relate the work done by the gas to the change in volume of cylinder A:

W = P_A ΔV = P_A (V_A - V_f)

where ΔV = change in volume of cylinder A, V_f = final volume of cylinder A.

Now, using Hooke's law, we can relate the change in volume of cylinder A to the distance the piston has moved:

ΔV = V_A - V_f = V_A - (V_A - x) = x

where x = distance the piston has moved.

Plugging this into our work equation, we get:

W = P_A x

Setting this equal to our work-energy equation,