Ideal gas question with 3 variables (scary)

[jrklx250s]

Homework Statement

A horizontal, insulated cylinder contains a frictionless non conducting piston. On each side of the piston is 54L of and inert monatomic ideal gas at 1 atm and 273K. Heat is slowly applied to the gas on the left side until the piston has compressed the gas on the right to 7.59 atm.

a) How much work is done on the gas on the right side?
b) What is the final temperature of the gas on the right side?
c) What is the final temperature of the gas on the left side?
d) How much heat was added to the gas on the left side?

The Attempt at a Solution

My efforts...

a) Since this is a monatomic gas we can conclude that the ratio fo heat capacity is close to 5/3 so gamma = 5/3

Assuming that there is no heat transferred from left to right I can use the equation PV^Y = constant where Y = gamma

therefore I have PiVi^Y = PfVf^Y
Since I am given Pi = 1 atm and Vi = 54L and Pf = 7.59 I can rearrange for Vf...

Then I can subsititute all these values into a previous work equation i derived

W = (PfVf - PiVi)/(Y-1)
and with all known values solve for Work


OK so is this solution possible I believe I have gone wrong somewhere here

On to b)

Here I can just use the ideal gas formula PV=nRT

Assuming there is 1 mole of gas in each side I can just take the pressure and volume of the gas that I found in a) and use n= 1 and R= 0.0821

However I feel this is wrong as well...

On to c)

Here if I know the pressure and volume of the left side and the pressure volume and temperature of the right side I can just make use of the equation

PiVi/Ti = PfVf/Tf

Here I am not sure if this is right as well...

On to d)

Well since we need to calculate the heat added I must make use of the equation

Q = ΔU -W

Here we can subsitute the work in for the value found in a) if it is correct because this will be the work supplied to the right side from the left side...however I am lost in calculated ΔU is it just the change in temperatures times the heat capacity at constant pressure? Once again I am confused in calculating the ΔU because on each side the temperature volume and pressure are all changing...

Thank you very much for any help in this complicated problem.
From a very frustrated student :D.

 

[ehild]

The Attempt at a Solution

My efforts...

a) Since this is a monatomic gas we can conclude that the ratio fo heat capacity is close to 5/3 so gamma = 5/3

Assuming that there is no heat transferred from left to right I can use the equation PV^Y = constant where Y = gamma

therefore I have PiVi^Y = PfVf^Y
Since I am given Pi = 1 atm and Vi = 54L and Pf = 7.59 I can rearrange for Vf...
?
Then I can subsititute all these values into a previous work equation i derived

W = (PfVf - PiVi)/(Y-1)
and with all known values solve for Work
?

Your formulae are not shown. Have you written them at all?

ehild

 

[jrklx250s]

No I didnt show my formulas because I didn't think I was even on the right track for this question. I mean I know you must use PV=nRT, PiVi^γ=PfVf^γ, and P1V1/T1 = P2V2/T2 and for the last one Q = ΔU - W...im just unsure if these are appropriate for each question

 

[ehild]

Those are the formulae you need. Go ahead.

ehild

 

[LiLi1212]

Can you please explain how we calculate part d with the information we got from previous parts.I think for du = C_v*dT , we don't have a value for C_v.Thanks

 

[Chestermiller]

Part A looks OK. In part B, you can't just assume that there is 1 mole. From the initial conditions, you can calculate the number of moles on each side of the piston before the heating takes place. For part C, you know the final pressure on the left side, the final volume of the left side, and the number of moles on the left side. This is enough to calculate the final temperature of the left side. To do part D, you know that the work done on the total system is zero. Therefore, the heat added has to equal the sum of the changes in internal energy for the two compartments. You know the initial and final temperatures of the two compartments, and the number of moles in each compartment, so you can calculate the changes in internal energy.

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