If the p-v curve is given by pV=constant, what is the final pressure?

Question

A gas undergoes a process that starts at a pressure of 25 psia and an initial volume of 4ft(cubed). Work is done on the gas in the amount of 12 Btu.

Answer 1

I agree with civil's analysis except that the question states that "work is done on the gas", not by it, so
-9360 = 14400*ln(v2/v1)
-0.65 = ln(v2) - ln(v1)
v2 = e^(-0.65 + ln(v1)) = e^(-0.65 + ln(4)) = 2.089 ft^3
p2 = p1*v1/v2 = 14400/2.089 * (1/144) = 47.87 psia
This means that the work was done by the mechanical equivalent of 12 BTU, not by adding that amount of heat, which is consistent with the rough definition, in thermodynamics, of work as any interaction that is not heat.

Answer 2

isothermal process
convert pressure to the correct units
p1 = 25 lbf/in^2 * (12 in /1 foot)^2 = 3600 lbf/ft^2
from the ideal gas law
pv = nrt = constant = 3600*4 = 14400
p = nrt/v = 14400/v

work = intergral from (v1 to v2) P dV = int(v1,v2) nrt/V dV
work = int(v1,v2) 14400/V dV = 14400*ln(v2/v1)

1 BTU = 780 ft*lbf

work = 780*12 = 9360 ft*lbf

9360 = 14400*ln(v2/v1)

.65 = ln(v2) - ln(v1)

v2 = e^(.65 + ln(v1)) = e^(.65 + ln(4)) = 7.66 ft^3

from p1v1 = p2v2

p2 = p1*v1/v2 = 14400/7.66 * (1/144) = 13.1 psia