Find the work done by the gas for the given volume and pressure. Assume that the pressure is inversely proportional to the volume. A quantity of...
Given pressure is inversely proportional the the volume of the gas,
So, `p=k/V`
where p is the pressure and V is the volume.
Initial pressure= 1000
Initial volume `(V_0)` = 2
Final volume `(V_1)` = 3
Let's calculate k, `k=pV`
`k=(1000)(2)`
`k=2000`
Work done (W) =`int_(V_0)^(V_1)k/VdV`
`=int_2^3(2000)/VdV`
Take the constant out,
`=2000int_2^3(dV)/V`
Use the common integral :`intdx/x=ln|x|`
`=2000[ln|x|]_2^3`
`=2000[ln(3)-ln(2)]`
`=2000(1.098612289-0.69314718)`
`=2000(0.405465108)`
`=810.9302162`
Work done by the gas `~~810.93` foot-pounds.
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