Question

An ideal substance obeys the relation pV=100 where V is in ft^3 and p is in Psia. Evaluate the reversible non- flow work done on or by the substance as the pressure increases from 10 psia to 100 psia.

Answer 1

Answer:

$W=-230.25\ psia.ft^3$

Explanation:

Given that

p V = 100

This is the isothermal process.

We know that non-flow work given as

$W=\int pdV$

$p_1=10\ psia$

$p_2=100\ psia$

We know that work for isothermal process  

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

For isothermal process$P_1V_1=P_2V_2=C$

Here C= 100

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

$W=100\ln \dfrac{10}{100}$

$W=-230.25\ psia.ft^3$

Negative sign indicates that work in done on the system.