Adiabatic expansion in a thermodynamic system

[cirimus]

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 :
p_1 = 1 atm = 1.013*10^5 N/m^2
and :
T_1 = 300 K

2) Isochoric proces resulting in :
p_2 = 3.p_1

3) Adiabatic expansion resulting in:
p_3 = p_1

4) Isobar compression resulting in
V_4 = V_1

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 :

pV = nRT
\frac{p_xV_x}{T_x} = \frac{p_y V_y} {T_y}
\Delta U = n Cv \Delta T

ideal gas + adiabatic expansion :
pV^{\lambda} = c

Note : lambda is Youngs module, c is a constant value.

diatomic :
C_v = 5/2 R
\lambda = 1.4

The Attempt at a Solution

b) I'm starting with this one since it seemed easier :
\frac{p_1 V_1}{T_1} = \frac{p_2 V_2} { T_2 }
\Rightarrow T_2 = \frac{p_2 V_2 T_1}{p_1 V_1}
\Rightarrow T_2 = 900
My guess is that my teacher approached this through using the
pV = nRT 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 :
p_2V_2^\lambda = constant = p_3V_3^\lambda
\Rightarrow V_3 = \log_\lambda ( \frac{p_2}{p_3} * V_2^\lambda)
\Rightarrow V_3 = \log_{1.4} ( 3 (4*10^{-3})^{1.4} )
\Rightarrow V_3 = -19.70870781

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 :

\sum W = W_{1,2} + W_{2,3} + W_{3,1}
\sum W = 0 + W_{2,3} + p_3 (V_1 - V_3)

But I'll have to integrate over an unknown volume to get the value of W_{2,3} ...
If I know the value of V_3 I think I can calculate the value of T_3 using p_3V_3=nRT_3. Once I know T_3 I can use \Delta U = n Cv \Delta T_{2,3} = - W_{2,3}.

 

[Andrew Mason]

b) the final temperature has to be 900 K.

a) Use the adiabatic condition PV^\gamma = K to give:

\left(\frac{V_f}{V_i}\right)^\gamma = \frac{P_i}{P_f}

c) The work done from 1-2 and 3-4 is easy. To determine the work done from 2-3 use the first law: dQ = dU + dW to determine the work done (what is Q for this adiabatic expansion?). Hint: you just have to know the change in temperature - use:

T_2V_2^{\gamma -1} = T_3V_3^{\gamma -1} to find the temperature at 3.)

AM

 

[cirimus]

I am using that formula for (a) in my attempted solution, but te result is not correct. The values I use are:

V_f = unknown
V_i = 4*10^{-3}
P_i = 3 * P_f
P_f = 1.013 * 10^{5}
\lambda = 1.40

Am i using wrong values ?

Attempt 2 :

V_f = V_i * \log_{1.4}(p_i / p_f)
\Rightarrow V_f = 4*10^{-3} * \log_{1.4}(3)
\Rightarrow V_f = 13 * 10^{-3}

 

[Andrew Mason]

I am using that formula for (a) in my attempted solution, but te result is not correct. The values I use are:

V_f = unknown
V_i = 4*10^{-3}
P_i = 3 * P_f
P_f = 1.013 * 10^{5}
\lambda = 1.40

Am i using wrong values ?

Attempt 2 :

V_f = V_i * \log_{1.4}(p_i / p_f)
\Rightarrow V_f = 4*10^{-3} * \log{1.4}(3)
\Rightarrow V_f = 13 * 10^{-3}

x^\gamma = (e^{\ln{x})^\gamma} = e^{\gamma\ln{x}AM

 

[Andrew Mason]

As a follow-up to my last post, in case you found it too cryptic: your figures are correct (although you should state the units, particularly in your answer). The problem is with algebra.

AM

 

[cirimus]

Thank you ! Using this and your previous posts I was able to find the answer for both (a) and (c) now. (how do i mark this thread as solved ?)