PROPERTIES OF MATTER BYU 12

Solution 12: Molar Specific Heat of Gas in a Soap Bubble

The molar specific heat $C$ is defined by $dQ = nCdT$. Using the first law: $dQ = dU + dW$. $$ nCdT = nC_v dT + P dV $$ Dividing by $n dT$: $$ C = C_v + \frac{P dV}{n dT} $$

For an ideal gas, $PV = nRT$. Differentiating both sides: $$ P dV + V dP = nR dT $$ $$ n dT = \frac{P dV + V dP}{R} $$ Substituting this into the expression for $C$: $$ C = C_v + \frac{P dV}{\frac{1}{R}(P dV + V dP)} = C_v + R \left( \frac{P dV}{P dV + V dP} \right) $$

The pressure inside the bubble is $P = P_0 + \frac{4\sigma}{r}$ and Volume $V = \frac{4}{3}\pi r^3$.
Differentiating with respect to $r$: $$ dP = -\frac{4\sigma}{r^2} dr $$ $$ dV = 4\pi r^2 dr $$

Now, calculate the ratio $\frac{V dP}{P dV}$: $$ \frac{V dP}{P dV} = \frac{(\frac{4}{3}\pi r^3)(-\frac{4\sigma}{r^2} dr)}{(P_0 + \frac{4\sigma}{r})(4\pi r^2 dr)} $$ $$ \frac{V dP}{P dV} = \frac{-\frac{16}{3}\pi \sigma r}{4\pi r^2(P_0 + \frac{4\sigma}{r})} = \frac{-\frac{4}{3}\sigma}{r(P_0 + \frac{4\sigma}{r})} = \frac{-4\sigma}{3P_0 r + 12\sigma} $$

$$ C = C_v + R \left( \frac{1}{1 + \frac{V dP}{P dV}} \right) = C_v + R \left( \frac{1}{1 – \frac{4\sigma}{3P_0 r + 12\sigma}} \right) $$ $$ C = C_v + R \left( \frac{3P_0 r + 12\sigma}{3P_0 r + 12\sigma – 4\sigma} \right) = C_v + R \left( \frac{3P_0 r + 12\sigma}{3P_0 r + 8\sigma} \right) $$

We know $C_v = C_p – R$. So: $$ C = (C_p – R) + R \left( \frac{3P_0 r + 12\sigma}{3P_0 r + 8\sigma} \right) $$ $$ C = C_p + R \left( \frac{3P_0 r + 12\sigma}{3P_0 r + 8\sigma} – 1 \right) $$ $$ C = C_p + R \left( \frac{3P_0 r + 12\sigma – (3P_0 r + 8\sigma)}{3P_0 r + 8\sigma} \right) $$