# Homework Help: Adiabatic expansion in a thermodynamic system

1. May 10, 2010

### cirimus

1. The problem statement, all variables and given/known data
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)

2. Relevant 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 :
$$C_v = 5/2 R$$
$$\lambda = 1.4$$

3. 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 $$\Delta U = n Cv \Delta T_{2,3} = - W_{2,3}$$.

Last edited: May 11, 2010
2. May 10, 2010

### 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

3. May 11, 2010

### 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 :

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

Last edited: May 11, 2010
4. May 11, 2010

### Andrew Mason

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

AM

5. May 11, 2010

### 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

6. May 11, 2010

### 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 ?)

Last edited: May 11, 2010