Answer to Question #128832 in Mechanics | Relativity for Ariyo Emmanuel

Question

1 kg of a fluid is compressed reversibly according to a law pv = 0.25, where p is in bar and v is in m3/kg. The final volume is one-fourth of the initial volume. Calculate the work done on the fluid and sketch the process on a p-v diagram.

Answer 1

The work that it done by fluid is an integral under the curve pV=0.25 in (p,V) coordinates. According to the task, the fluid is compressed, so the work is done on fluid (has minus sign):

$W = -\int_{V_1}^{V_2} pdV =\int_{V_2}^{V_1} pdV$

from a low of fluid compression we can find $pV = 0.25 \Rightarrow p = \frac{0.25}{V}*10^5$ (Pa). The multiplier $10^5$ appeared because we have converted bar to Pa (SI units), 1 bar = $10^5$ Pa. Also, we know that $V_2 = 0.25 V_1$ . Now we can calculate the integral:

$W = \int_{0.25V_1}^{V_1} \frac{0.25*10^5}{V}dV = 0.25*10^5 ln (\frac{V_1}{0.25V_1})=0.25*10^5 ln4 = 34 657$ J $\approx 34.7$ kJ