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 fr…

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asked Dec 3, 2020 102k views

1 Answer

Pointy · Answer 1 · 2020-12-06T23:24:11+0000

Answer:

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

Step-by-step 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

$W=P_1V_1\ln (P_1)/(P_2)$

For isothermal process
P_1V_1=P_2V_2=C

Here C= 100

$W=P_1V_1\ln (P_1)/(P_2)$

$W=100\ln (10)/(100)$

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

Negative sign indicates that work in done on the system.