Question

The buoyancy force acting on the gas bubble is (Assume R is the universal gas constant):

• A. ${\rho }_{\ell }nRg{T}_0\frac{(P_0+{\rho }_{\ell }gH{)}^{2/5}}{(P_0+{\rho }_{\ell }gy{)}^{7/5}}$
• B. $\frac{{\rho }_{\ell }nRg{T}_0}{(P_0+{\rho }_{\ell }gH{)}^{2/5}[P_0+{\rho }_{\ell }g(H-y){]}^{3/5}}$
• C. ${\rho }_{\ell }nRg{T}_0\frac{(P_0+{\rho }_{\ell }gH{)}^{3/5}}{(P_0+{\rho }_{\ell }gy{)}^{8/5}}$
• D. $\frac{{\rho }_{\ell }nRg{T}_0}{(P_0+{\rho }_{\ell }gH{)}^{3/5}[P_0+{\rho }_{\ell }g(H-y){]}^{2/5}}$

Answer 1

The correct option is B $\frac{{\rho }_{\ell }nRg{T}_0}{(P_0+{\rho }_{\ell }gH{)}^{2/5}[P_0+{\rho }_{\ell }g(H-y){]}^{3/5}}$

As there is no exchange of heat, the process is adaibatic.
For an adiabatic process, $T{P}^{\frac{1-\gamma }{\gamma }}=constant$
$\Rightarrow T_1{P}_1^{\frac{1-\gamma }{\gamma }}=T_2{P}_2^{\frac{1-\gamma }{\gamma }}$
Substituting,
$P_1=P_0+{\rho }_{l}gH$ and $P_2=P_0+{\rho }_{l}g(H-y)$ and $\gamma =\frac{5}{3}$ and $\frac{1-\gamma }{\gamma }=\frac{1-\frac{5}{3}}{\frac{5}{3}}=-\frac{2}{5}$
$T_2=T_1(\frac{P_1}{P_2}{)}^{-\frac{2}{5}}=T_1(\frac{P_2}{P_1}{)}^{\frac{2}{5}}=T_0(\frac{P_0+{\rho }_{l}g(H-y)}{P_0+{\rho }_{l}gH}{)}^{\frac{2}{5}}$
Buoyancy force $F_b={\rho }_{l}vg={\rho }_{l}g\left(\frac{nR{T}_2}{P_2}\right)$
Substituting $T_2$ and $P_2$
we get $F_b=\frac{{\rho }_{\ell }nRg{T}_0}{(P_0+{\rho }_{\ell }gH{)}^{2/5}[P_0+{\rho }_{\ell }g(H-y){]}^{3/5}}$