Posted this in the wrong place, please help with this physics problem :( im so stressed?

Question

An expandable cylinder has its top connected to a spring with force constant 2.00 103 N/m. (See Fig. P10.58.) The cylinder is filled with 4.50 L of gas with the spring relaxed at a pressure of 1.00 atm and a temperature of 20.0°C.

Here is the link to the figure:

http://i136.photobucket.com/albums/q200/physics2006/physics2.gif

ANd the questions are:

a) If the lid has a cross-sectional area of 0.0100 m2 and negligible mass, how high will the lid rise when the temperature is raised to T = 260°C?

cm

(b) What is the pressure of the gas at 260°C?

Pa

Answer 1

P1V1/T1 = P2V2/T2
P1 = 1 atmosphere = 101325 N / (m^2)
V1 = (4*10^3 cm^3)(1 m^3/10^6 cm^3) = 4*10^-3 m^3
P2 = P1 + h*2.00103/0.01
V2 = V1 + 0.01h
T1 = 20°C = 293.15°K
T2 = 260°C + 273.15 = 533.15°K
A = 0.01 m^2
P1V1/T1 = (P1 + hk)(V1 + Ah)/T2
P1V1/T1 = P1V1/T2 + ((kV1 + AP1)h + kAh^2)/T2
P1V1(T2 - T1)/T1 = (kV1 + AP1)h + kAh^2
kAh^2 + (kV1 + AP1)h - P1V1(T2 - T1)/T1
h^2 + ((kV1 + AP1)/kA)h - P1V1(T2 - T1)/(kAT1)
((kV1 + AP1)/kA) = ((2.00103*4*10^-3 + 0.01*101325))/(2.00103*0.01)
((kV1 + AP1)/kA) = 50636.82
P1V1(T2 - T1)/(kAT1) = (101325)(4*10^-3)(533.15 - 293.15)/(2.00103*0.01*293.15)
P1V1(T2 - T1)/(kAT1) = 16582.28
h = (- 50636.82 ± √(50636.82^2 + 4*16582.28))/2
h = - 25318.41 ± √(25318.41^2 + 16582.28)
h = - 25318.41 ± 25318.74
h = 0.34 m = 340 cm

b)
P2 = 101,325 + 0.34*2.00103/0.01
P2 = 101,393 N/m^2

Answer 2

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