PROPERTIES OF MATTER ChYU 3

1. Potential Energy of the Grid

Consider a square grid of side $L$ made of springs of initial length $a$.

• Number of springs along one side: $L_0/a$.
• Total number of springs (for large grid): $N \approx 2 \times (L_0/a)^2$.
• Stretched length of each spring: $x = a (L/L_0)$.
• Extension of each spring: $\delta = x – a = a(\frac{L}{L_0} – 1)$.

Total Potential Energy ($U$):

$$ U = N \times \frac{1}{2} k \delta^2 \approx 2 \left( \frac{L_0}{a} \right)^2 \times \frac{1}{2} k \left[ a \left( \frac{L}{L_0} – 1 \right) \right]^2 $$ $$ U = \frac{L_0^2}{a^2} k a^2 \frac{(L – L_0)^2}{L_0^2} = k (L – L_0)^2 $$

Since Area $S = L^2$, we have $L = \sqrt{S}$ and $L_0 = \sqrt{S_0}$.

$$ U = k (\sqrt{S} – \sqrt{S_0})^2 $$

2. Pressure inside Spherical Balloon

For a sphere, $S = 4\pi r^2$ and $S_0 = 4\pi r_0^2$. Substitute into the energy equation:

$$ U = k (\sqrt{4\pi r^2} – \sqrt{4\pi r_0^2})^2 = 4\pi k (r – r_0)^2 $$

The work done by the internal pressure $P$ in expanding the balloon volume by $dV$ is equal to the increase in potential energy $dU$ (Principle of Virtual Work).

$$ dW = P \, dV = dU $$

Volume $V = \frac{4}{3}\pi r^3 \implies dV = 4\pi r^2 dr$.

Differentiating Energy $U$ with respect to $r$:

$$ dU = \frac{d}{dr} [ 4\pi k (r – r_0)^2 ] dr = 8\pi k (r – r_0) dr $$

Equating $P \, dV = dU$:

$$ P (4\pi r^2 dr) = 8\pi k (r – r_0) dr $$ $$ P = \frac{8\pi k (r – r_0)}{4\pi r^2} $$