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cirimus
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Homework Statement
4 liters of an ideal diatomic gas are compressed in a cilinder. In a closed process, the following steps are taken :
1) Initial state :
[tex]p_1 = 1 atm = 1.013*10^5 N/m^2 [/tex]
and :
[tex]T_1 = 300 K[/tex]
2) Isochoric proces resulting in :
[tex]p_2 = 3.p_1[/tex]
3) Adiabatic expansion resulting in:
[tex]p_3 = p_1[/tex]
4) Isobar compression resulting in
[tex] V_4 = V_1 [/tex]
Note : State 4 = State 1
Questions :
a) What is the volume of the gas at the end of the adiabatic proces ? (solution : 8.77*10^-3 )
b) What is the temperature of just before the adiabatic expansion? (solution : 902 K)
c) What is the work performed by the gas in this cycle ? (solution : 335.0 J)
Homework Equations
ideal gas :
[tex] pV = nRT [/tex]
[tex] \frac{p_xV_x}{T_x} = \frac{p_y V_y} {T_y} [/tex]
[tex]\Delta U = n Cv \Delta T[/tex]
ideal gas + adiabatic expansion :
[tex]pV^{\lambda} = c[/tex]
Note : lambda is Youngs module, c is a constant value.
diatomic :
[tex]C_v = 5/2 R[/tex]
[tex]\lambda = 1.4[/tex]
The Attempt at a Solution
b) I'm starting with this one since it seemed easier :
[tex]\frac{p_1 V_1}{T_1} = \frac{p_2 V_2} { T_2 }[/tex]
[tex]\Rightarrow T_2 = \frac{p_2 V_2 T_1}{p_1 V_1}[/tex]
[tex]\Rightarrow T_2 = 900 [/tex]
My guess is that my teacher approached this through using the
[tex]pV = nRT[/tex] equation twice, introducing rounding errors ?
a) I've tried to approach this in several ways, but none seem to give me the correct solution... My current approach :
Since we know it's an ideal gas that undergoes an adiabatic expansion, we could use the formula given above :
[tex] p_2V_2^\lambda = constant = p_3V_3^\lambda [/tex]
[tex] \Rightarrow V_3 = \log_\lambda ( \frac{p_2}{p_3} * V_2^\lambda)[/tex]
[tex] \Rightarrow V_3 = \log_{1.4} ( 3 (4*10^{-3})^{1.4} ) [/tex]
[tex] \Rightarrow V_3 = -19.70870781 [/tex]
That is wrong in many ways, but I don't know which assumption I made is wrong ...
c) Don't know yet, I'm guessing something like :
[tex] \sum W = W_{1,2} + W_{2,3} + W_{3,1}[/tex]
[tex] \sum W = 0 + W_{2,3} + p_3 (V_1 - V_3) [/tex]
But I'll have to integrate over an unknown volume to get the value of [tex]W_{2,3}[/tex] ...
If I know the value of [tex]V_3[/tex] I think I can calculate the value of [tex]T_3[/tex] using [tex]p_3V_3=nRT_3[/tex]. Once I know [tex]T_3[/tex] I can use [tex]\Delta U = n Cv \Delta T_{2,3} = - W_{2,3}[/tex].
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