integration word problem help

**Tweety** · November 29th 2010, 12:47 PM

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

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

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

Calculate Wrev when

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

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

c) $P = \frac{RT}{(V-b)} - \frac{a}{V^{2}}$ - 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?

$PV =RT$

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

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

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

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

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

is this correct?

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

Thank you.

**SammyS** · November 29th 2010, 06:56 PM

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?

$PV =RT$

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

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

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

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

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

is this correct?

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

Thank you.

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

Multiply your answer by -RT.

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

$W_{rev}=-\int_{V_1}^{V_2} KV^{-\gamma} d V$ , 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:

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

I expect that you can do these integrations also.

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