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Physics work done question!!!1

A unit mass of fluid is compressed reversibly according to the law pv=0.5, where p is the pressure in bar and v is the specific volume in m3/kg. The final volume is ¼ of the initial volume. Calculate the work done on the fluid.
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Avatar for jaroc
jaroc
6
Original post by jas248
A unit mass of fluid is compressed reversibly according to the law pv=0.5, where p is the pressure in bar and v is the specific volume in m3/kg. The final volume is ¼ of the initial volume. Calculate the work done on the fluid.


This should help:

The work done is the area under the pressure-volume graph so that

W=∫V1V2 ⁣ ⁣p  dV\displaystyle W=\int_{V_1}^{V_2}\! \! p\; \mathrm{d}V.

If you still can't do it, let me know.
Reply 2
Avatar for jas248
jas248
OP
Original post by jaroc
This should help:

The work done is the area under the pressure-volume graph so that

W=∫V1V2 ⁣ ⁣p  dV\displaystyle W=\int_{V_1}^{V_2}\! \! p\; \mathrm{d}V.

If you still can't do it, let me know.


Thanks for the reply but im really not that great with intergration however I did find another formula:

W=PVln(v1/v2)
Therefore W=0.5*10^5 ln(1/0.25)
=69314.7 J

Am i right in using this formula????

Thanks for your help
Reply 3
Avatar for jaroc
jaroc
6
Original post by jas248
Thanks for the reply but im really not that great with intergration however I did find another formula:

W=PVln(v1/v2)
Therefore W=0.5*10^5 ln(1/0.25)
=69314.7 J

Am i right in using this formula????

Thanks for your help


Yeah, that's correct. This is the formula for isothermal process, i.e. where temperature is constant. According to the Clapeyron's formula, if temperature is constant, pV is also constant. So this an isothermal process. Below is the derivation of this formula:

Spoiler

Reply 4
Avatar for sumit95
sumit95
Can someone explain what it means by "unit mass of fluid" what is a unit mass is it a slug (14.6kg)? Because dont you have to convert the m3/kg into m3?

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