# Work on expanding gasses

1. Nov 7, 2005

### dimensionless

I've been trying to tackle the following problem, but I can't seem to get it right.
An ideal monatomic gas expands quasi-statically to twice its volume. If the process is isothermal, the work done by the gas is W_i. If the process is adiabatic, the work doen by the gas is W_a. Show that 0 < W_a < W_i.
For isothermal processes I have
$$W_i = \int_{v_i}^{v_f} \frac{nRT}{V}dV = nRT ln(2)$$
$$W_a = k \int_{v_i}^{v_f} \frac{dV}{V^{\gamma}}$$
Where do go from this point?

2. Nov 7, 2005

### Andrew Mason

What is the antiderivative of:

$$\frac{1}{V^{\gamma}} = V^{-\gamma}$$?

Also: $k = P_iV_i^\gamma$

What is the maximum value of $\gamma$ ?

AM

3. Nov 7, 2005

### dimensionless

I've got $$\gamma = C_p/C_v$$ written down as the only definition for $$\gamma$$.

I don't know of any alternative for $$V^{-\gamma}$$.

4. Nov 7, 2005

### Andrew Mason

But what is the maximum value it can have? What is the minimum value of $C_v$? What is $C_p - C_v$?

That's ANTI-DERIVATIVE. What is the function F(x) such that $F'(x) = x^n$?

AM

5. Nov 8, 2005

### dimensionless

I've got $$C_p - C_v = \frac{5}{2}R-\frac{3}{2}R=R$$
and $$V^{-\gamma} = -\gamma V^{-\gamma-1} dV$$

I've worked this out but I get $$0 < W_i < W_a$$ instead of $$0 < W_a < W_i$$ for an answer.

Last edited: Nov 8, 2005
6. Nov 8, 2005

### Andrew Mason

?? How do you figure that?

The antiderivative of $V^{-\gamma}$ is: $\frac{1}{-\gamma +1}V^{-\gamma + 1}$

$$\int_{V_i}^{V_f}\frac{dV}{V^\gamma} = K\left(\frac{V_f^{-\gamma + 1} - V_i^{-\gamma + 1}}{-\gamma + 1}\right)$$

Use the fact that the minimum value of C_v = 3R/2 and $-\gamma + 1 = 1 - \gamma = -(C_p - C_v)/C_v$ to work out the inequality.

AM

Last edited: Nov 8, 2005
7. Nov 8, 2005

### Dr.Brain

Do this:

For the first process , the isothermal one , we know for an ideal gas :

$PV = K$

Draw the Pressure versus Volume diagram for the first process (isothermal) , with values as per given in the question.

------------------------------------------------------------------------

For the second process , draw the Pressure versus Volume diagram , you know that for an adibatic change:

$PV^(\gamma) = K$

Take P_1 and P_2 for both the processes to be the same.

Because you know area under PV graph gives the work done. You would find that because of the factor "gamma" for adiabatic process , it will have a steeper slope in the graph and thus lesser area as compared to that of isothermal process.

BJ