A bubble has surface tension S. The ideal gas | vedclass

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Question

(D) For a spherical bubble,the pressure inside is $P_{gas} = P_a + \frac{4S}{r}$.If the surface is a perfect heat insulator,the process is adiabatic:

Vedclass · Accepted Answer

$B, D$

Vedclass · Answer

(D) For a spherical bubble,the pressure inside is $P_{gas} = P_a + \frac{4S}{r}$.If the surface is a perfect heat insulator,the process is adiabatic: $PV^{\gamma} = 	ext{constant}$.Since $V = \frac{4}{3}\pi r^3$,we have $\left(P_a + \frac{4S}{r}ight) (r^3)^{5/3} = 	ext{constant}$,which implies $\left(P_a + \frac{4S}{r}ight) r^5 = 	ext{constant}$.Thus,$\left(P_{a1} + \frac{4S}{r_1}ight) r_1^5 = \left(P_{a2} + \frac{4S}{r_2}ight) r_2^5$,or $\left(\frac{r_1}{r_2}ight)^5 = \frac{P_{a2} + \frac{4S}{r_2}}{P_{a1} + \frac{4S}{r_1}}$. So,$(A)$ is correct.For an adiabatic process,$P^{1-\gamma} T^{\gamma} = 	ext{constant}$.Substituting $P = P_a + \frac{4S}{r}$ and $\gamma = 5/3$,we get $\left(P_a + \frac{4S}{r}ight)^{-2/3} T^{5/3} = 	ext{constant}$.This leads to $\left(\frac{T_2}{T_1}ight)^{5/3} = \left(\frac{P_{a2} + \frac{4S}{r_2}}{P_{a1} + \frac{4S}{r_1}}ight)^{2/3}$,which simplifies to $\left(\frac{T_2}{T_1}ight)^{5/2} = \frac{P_{a2} + \frac{4S}{r_2}}{P_{a1} + \frac{4S}{r_1}}$. So,$(D)$ is correct.If the surface is a perfect heat conductor and temperature is constant,$PV = 	ext{constant}$,so $\left(P_{a1} + \frac{4S}{r_1}ight) r_1^3 = \left(P_{a2} + \frac{4S}{r_2}ight) r_2^3$,which contradicts $(C)$.Therefore,$(A)$ and $(D)$ are correct.

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