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
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 im 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 Im 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 im lost in calculated ΔU is it just the change in temperatures times the heat capacity at constant pressure? Once again Im 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.