Answer
a) $P_2 = 188.31\ psi$
b) $W = 281.75\ BTU$
c) $f = 85.5\%$
Work Step by Step
a)From the information given, The elastic force equation is given by:
$F = k.\Delta x$ ($F\propto x$, and initially touches the piston but exerts no force on it)
The final volume is $V_2=2V_1 = 30ft³$
The displacement is: $\Delta x = \frac{\Delta V}{A} = 5\ ft$
The pressure increase $\Delta P = \frac{F}{A} = \frac{k. \Delta x}{A} = 25000\ psf = 173.61\ psi $
The final pressure is: $P_2 = 188.31\ psi$
b) Work done is the area of the trapezoid defined by the points $(P_1,V_1), (P_2,V_2)$ on the P-V diagram:
$W = (V_2-V_1)(\frac{P_2+P_1}{2})$
$W = 1552.58\ psi.ft³ \times \frac{1\ BTU}{5.40395\ psi.ft³} = 281.75\ BTU$
c) The work done on the spring is:
$W _s= \frac{k.\Delta x^2}{2}$
$W_s = 187500\ lbf.ft \times \frac{1\ BTU}{778.17\ lbf.ft} = 240.95\ BTU$
The fraction is given by:
$f = \frac{W_s}{W}\times 100\% $
$f = 85.5\%$
