A soap bubble of radius r contains a monoatom | vedclass

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(B) For a soap bubble,the excess pressure is $P = \frac{4T}{r}$,where $T$ is the surface tension. Since atmospheric pressure is neglected,the pressure

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(B) For a soap bubble,the excess pressure is $P = \frac{4T}{r}$,where $T$ is the surface tension. Since atmospheric pressure is neglected,the pressure of the gas is $P = \frac{4T}{r}$.From the ideal gas law,$PV = nRT$. Substituting $P$,we get $\left(\frac{4T}{r}\right) \left(\frac{4}{3}\pi r^3\right) = nRT$,which simplifies to $\frac{16\pi T r^2}{3} = nRT$.Since $T$ (surface tension) is constant,$n R dT = d(\frac{16\pi T r^2}{3}) = \frac{32\pi T r dr}{3}$.The heat supplied is $dQ = dU + dW = \frac{nRdT}{\gamma-1} + PdV$.For a monoatomic gas,$\gamma = 5/3$,so $\frac{1}{\gamma-1} = 1.5 = 3/2$.$dQ = \frac{3}{2} nRdT + PdV$.$dV = d(\frac{4}{3}\pi r^3) = 4\pi r^2 dr$.$PdV = (\frac{4T}{r})(4\pi r^2 dr) = 16\pi T r dr$.From $n R dT = \frac{32\pi T r dr}{3}$,we have $16\pi T r dr = \frac{3}{2} nRdT$.Substituting this into the heat equation: $dQ = \frac{3}{2} nRdT + \frac{3}{2} nRdT = 3nRdT$.The molar heat capacity $C = \frac{dQ}{ndT} = 3R$.

$A$. Isobaric	$I$. $\Delta Q = \Delta W$
$B$. Isochoric	$II$. $\Delta Q = \Delta U$
$C$. Adiabatic	$III$. $\Delta Q = 0$
$D$. Isothermal	$IV$. $\Delta Q = \Delta U + P \Delta V$

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