I need help with this problem (except the first part, which I solved correctly).

Question

I need help with this problem (except the first part, which I solved correctly). A gas container had a piston which could move smoothly olong the container wall. Two identical springs were attached onto the outer surface of the piston. The springs in their natural lengths touched a ring which was locked onto the container wall. Heat energy was absorbed by the gas and the gas expanded by moving the piston and compressing the springs. Then the locking mechanism of the ring was released such that the ring became a projectile and the springs returned to their natural lengths Then an external force moved the piston and the springs back to the initial position. The cycle in the PV diagram wauld be A to B to C and then back to A All of the above descriptions can be summarized in a PV Diagram One mole 6.023623 particles, Boltzmann k 1.38E-23 Joule per Kelvin, One ATM 1.01E5 N/sq meter Nobel gases such as helium, neon can be modeled as monatomic ideal gas with Cp 20.8 J/K/mol and Cv 12.5 J/K/mol. A small mixture with diatomic gas such as oxygen would increase Cp and Cv values Q1: A One ATM 2 moles helium gas 25 Celsius. Find the volume V: 0.05m 02: B- 1.8 ATM from heating of gas. Find the heat energy in the A to B process, assuming a linear path inthe PV diagram, (Q. U-change + w-change), a,dVrgw ??-n(gar Q3: If the heat source came from friction of 12 Newton, find the distapce required Q4: C= One ATM. Find the volume, assuming a reversible quasi-adiabatic process for the launch, and the work done by the gas while reducing the gas speed-10% of the work done from A to B process. Q5: For the A to C process, find the work done by the external force on the gas and the internal energy loss of the gas as heat dumped onto the environment 06: For the A to B to C process, find the sum of the work by the gas and the wort on gas

Explanation / Answer

1. Pa = 1 atm

n = 2 moles

Ta = 25 C

hence

PaVa = nRTa

Va = nRTa/Pa = 0.0490635564 m^3

1 atm = 1.01*10^5 Pa

Ta = 25 + 273.16 K

2. Pa = 1.8 atm

for alinear path

P and V are proportional

P/V is contant = k

hence

Pa/Va = Pb/Vb

Vb = 0.08831440152 m^3

heat energy = Q

internal energy increase = U

hence

Q = U + W

dW = PdV = kVdV

W = (Pa/Va)(Vb^2 - Va^2)/2 = 0.054951183168 atm m^3 = 5550.069499968 J

U = nCv(Tb - Ta) = nCv(PbVb/nR - PaVa/nR) ) = Cv(PbVb - PaVa)/R

Cv = 3R/2

hecne

U = 3(PbVb - PaVa)/2 = 16650.20849 J

hence

Q = 22200.277999872 J

3. d = ?

F = 12 N

hence

12*d = 22200.277999872

d = 1850.02316665 m

4. Pc = 1 atm

adiabatic process

work done by gas while reducing gas speed = 0.1*5550.069499968 J

hence

change in internal energy = -5550.069499968*0.1 J = 3(PcVc - PbVb)/2

Vc = 0.1553025105248 m^3

5. work done by external force = 5550.069499968 + 0.1*5550.069499968= 6105.0764499648 J

internal energy loss = -6105.0764499648J

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