Work done in isothermal expansion from 40 to 10 psig

[shawn100]

If air has a pressure of 40 psig and a volume of 8 cu. ft. expands isothermally to a pressure of 10 psig, find the external work performed during the expansion. How do I do this, do I first have to change 40 psig to psia, and how do I do that? This question has me lost! Any help appreciated, even a formula for me to understand it would help. thank you

 

[Andrew Mason]

If air has a pressure of 40 psig and a volume of 8 cu. ft. expands isothermally to a pressure of 10 psig, find the external work performed during the expansion. How do I do this, do I first have to change 40 psig to psia, and how do I do that? This question has me lost! Any help appreciated, even a formula for me to understand it would help. thank you

Use PV=nRT.

If T is constant then $P_iV_i = P_fV_f$. So you can work out what the final volume is.

The work is:

$$W = \int_{P_i}^{P_f} PdV = \int_{V_i}^{V_f} \frac{nRT}{V}dV$$

You have to work out that integral (hint: $\frac{d}{dV}ln V = 1/V$) and plug in the initial and final volumes.

You don't have to do any conversions. You just need the pressure ratio P_f/P_i.

AM

[edit: this last comment is not quite correct. You do have to work out nRT = P_iV_i which means you have to do a conversion. PSIA is absolute pressure in pounds/in^2, which means you have to include atmospheric pressure. PSIG is gauge pressure, which is 1 atm less than actual. It is easier to work in MKS. I would convert to MKS and then convert back.]

AM