Question

Carbon dioxide contained in a piston–cylinder device is compressed from 0.3 to 0.1 m3. During the process, the pressure and volume are related by P = aV–2, where a = 6.5 kPa·m6.
Calculate the work done on carbon dioxide during this process.

Answer 1

Answer:

$W = -6.5 Kpa m^6 (\frac{1}{0.1 m^3}- \frac{1}{0.3 m^3})= -6.5 Kpa m^6 (6.66 m^-3) =-43.33 Kpa m^3 = -43.33 KJ$

Explanation:

For this case we know the following info :

$V_i = 0.3 m^3$ represent the initial volume

$V_f = 0.1 m^3$ represent the final volume

We know that the pressure and volume are related with the following expression:

$P = aV^{-2}= \frac{a}{V^2}$

Where a is a constant given $a = 6.5 Kpa m^6$

And we need to calculate the work associated to this process.

We have a compression, and by definition the work is defined with the following general expression:

$W = \int_{V_i}^{V_f} P dV$

If we replace the expression for P we got:

$W = \int_{V_i}^{V_f} a V^{-2} dV$

If we integrate we got:

$W = -a (\frac{1}{V}) \Big|_{V_i}^{V_f}$

Using the fundamental theorem of calculus we have:

$W = -a (\frac{1}{V_f} -\frac{1}{V_i})$

And replacing the values we got:

$W = -6.5 Kpa m^6 (\frac{1}{0.1 m^3}- \frac{1}{0.3 m^3})= -6.5 Kpa m^6 (6.66 m^-3) =-43.33 Kpa m^3 = -43.33 KJ$