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

`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.