${}_t=\frac{V}{V}kT(-\frac{1}{{}^2})=\frac{V}{V}kT(\frac{L}{L_0}-(\frac{L_0}{L}{)}^2)$

We have a rubber specimon which is elongated by

$40\%$ at $T=350k,($so $=1\mathrm{.}4)$

The measured stress ${}_t=0\mathrm{.}73{10}^6$ Pascals

$(1$ Pascal $=1\frac{(J)}{m^3}=1\frac{(N)}{m^2})$

a$)$ Using the expression for ${}_t$ at the bottom of p$.36$

evaluate the density of crosslinks

b$)$ if the crosslinks are equally spaced in a cubic arrangement,

so $a=$ crosslink$-$crosslint spacing

$=\frac{1}{(\frac{V}{V}{)}^{\frac{1}{3}}},(convince$ yours$\frac{e}{f})$

evaluate $a$ in $m$ and &

c$)$ at $10\%$ elongation $(=1\mathrm{.}1)$

${}_t=0\mathrm{.}23{10}^6$ pascals $atT=350K$

is this data consistent with the result of a$)?$