# Cylinder & Piston

## Homework Statement

A vertical cylinder contains 500 moles of a monatomic ideal gas and is
closed off by a piston of mass 50 kg and a cross-sectional area of 100 cm2. The whole system is thermally insulated during the entire process described below. Initially, the piston is clamped in position so that the gas occupies a volume of 1 m3 and is at a temperature of 300 K. The piston is then released, and eventually comes to rest in a final equilibrium position corresponding to some larger volume V of the gas. Neglect any frictional forces which might prevent the piston from sliding freely within the cylinder. Compute:

(a) the number of gas molecules in the cylinder;

(b) the initial pressure of the gas (in units of atm);

(c) the final pressure of the gas (in units of atm).

(d) Obtain an expression for the work done by the expanding gas in terms of change in volume of the system.

Using the above results, apply the first law of thermodynamics and compute:

(e) the final volume of the gas; and

(f) the final temperature of the gas.
[HINT: The final equilibrium position of the piston is reached when the weight of the piston is exactly balanced by the pressure of the gas. Note: the process described above is not reversible.]

## The Attempt at a Solution

......Part A) i believe ij ust multiply 500 moles by avagdrs nubers

N = 500(6.02E^23) = 3.01E^26

......Part B) P = NkT/V = (3.01E^23)(1.38E^-23)(300)/1 = 1246140 N/m^2

N/^2 = 9.86923E^-6 atm

Therefore P_o = 12.29844227 atm

......Part C)

P = F/A = 50(9.81)/.1 = 490500 N/m^2 = 4.840857 atm (i think this is wrong, does anything less than 1 atm imply there is a vacume)

......Part D) i believe i used W = integral of Pdv,

v_o = 1 ^3, v_final = Nkt/P = (3.01E^23)(1.38E^-23)(300)/490500 = .00254 m^3

the vfinal should be larger than v initial because the pressure went down. what did i do wrong

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ideasrule
Homework Helper

## Homework Statement

......Part A) i believe ij ust multiply 500 moles by avagdrs nubers

N = 500(6.02E^23) = 3.01E^26

......Part B) P = NkT/V = (3.01E^23)(1.38E^-23)(300)/1 = 1246140 N/m^2

N/^2 = 9.86923E^-6 atm

Therefore P_o = 12.29844227 atm
All of that is right.

......Part C)

P = F/A = 50(9.81)/.1 = 490500 N/m^2 = 4.840857 atm (i think this is wrong, does anything less than 1 atm imply there is a vacume)
You have to take into account the pressure exerted by the atmosphere. Also, 100 cm^2 is 0.01 m^2, not 0.1 m^2.

P = F/A = 50(9.81)/.01 = 49050 N/m^2 = .4840857 atm

Ok so that is the Pressure frm the weight of the piston

Do i have to add in 1 atm, making it 1.5 atm

ideasrule
Homework Helper
Yeah.

because it desnt matter whats in the cylinder or how much, it only depends on the applied forces, if it pushes down with x amount, the gas is going to compress and push back with x amount pressure.... is this ture

Part D)

is this siply W = integral PdV from v_o=1 to V = nRT/P

ideasrule
Homework Helper
because it desnt matter whats in the cylinder or how much, it only depends on the applied forces, if it pushes down with x amount, the gas is going to compress and push back with x amount pressure.... is this ture
I don't understand this. What's "it"? What's "x amount"?

ideasrule
Homework Helper
Part D)

is this siply W = integral PdV from v_o=1 to V = nRT/P
Yes, but you have to figure out how to integrate PdV

Do you mean integrating based on the type of process,

what do you mean i have to figure out how to integrate PdV,

ideasrule
Homework Helper
Forget about that for a moment. Just find the answer to d): what is the work done by the expanding gas?

W = integral PdV from v_o=1 to V = nRT/P

=P(nRT/P -1) = nRT - P