1. integration word problem help

The reversible work per mol done, Wrev, to expand a fuid from a molar volume
V1 to a molar volume V2 is given by

$\displaystyle W_{rev} = - \int^{V_{2}}_V_{{1}} P dV$

Where P the the pressure of the fluid at a volume V.

Calculate Wrev when

a) $\displaystyle PV = RT$ - - i.e. an ideal gas. Assume that the temperature, T, is constant
(i.e. an isothermal expansion).

b) $\displaystyle P = KV^{- \gamma}$ and K, $\displaystyle \gamma$ are constants. Adiabatic expansion of an ideal gas. For information's sake $\displaystyle \gamma = \frac{C_{p}}{C_{v}}$ where where Cp and Cv are the heat capacities at constant pressure and volume respectively.

c) $\displaystyle P = \frac{RT}{(V-b)} - \frac{a}{V^{2}}$ - van der Waals equation of state with a and b constants,
assume that the temperature T is constant.
I think I am able to do part 'a', can someone check my working and let me know if it is correct?

$\displaystyle PV =RT$

$\displaystyle P = \frac{RT}{V}$

$\displaystyle \int^{v_{2}}_v_{1} \frac{RT}{V} dv$

since R and T are constants it can be left out of the integration so;

$\displaystyle RT \int^{v_{2}}_v_{1}\frac{1}{V} dv$

= $\displaystyle ln(V)$ evaluating that I get $\displaystyle ln(V_{2})- ln(V_{1})$

is this correct?

Also need help with part 'b' and 'c' please.

Thank you.

2. Thermo- Reversible Work

Originally Posted by Tweety
I think I am able to do part 'a', can someone check my working and let me know if it is correct?

$\displaystyle PV =RT$

$\displaystyle P = \frac{RT}{V}$

$\displaystyle \int^{v_{2}}_{v_{1}} \frac{RT}{V} dV$

since R and T are constants it can be left out of the integration so;

$\displaystyle RT \int^{v_{2}}_{v_{1}}\frac{1}{V} dv$

= $\displaystyle ln(V)$ evaluating that I get $\displaystyle ln(V_{2})- ln(V_{1})$

is this correct?

Also need help with part 'b' and 'c' please.

Thank you.

You dropped the negative sign, then you dropped the RT.

Part (b);
Simply substitute $\displaystyle KV^{-\gamma}$ in for $\displaystyle P$.

$\displaystyle W_{rev}=-\int_{V_1}^{V_2} KV^{-\gamma} d V$, K is a constant, so it comes out of the integral. That's a fairly basic integral.

Part (c)
For a van der Vaal's gas:

$\displaystyle \displaystyle W_{rev}=-\int_{V_1}^{V_2} \left(\frac{RT}{(V-b)} - \frac{a}{V^{2}} \right) d V =-RT \int_{V_1}^{V_2} \frac{1}{(V-b)}\ dV +a \int_{V_1}^{V_2} \frac{1}{V^{2}}\ d V$

I expect that you can do these integrations also.