Question

When the gas bubble is at a height y from the bottom, its temperature is:

• A. $T_0(\frac{P_0+{\rho }_{l}gH}{P_0+{\rho }_{l}gy}{)}^{2/5}$
• B. $T_0(\frac{P_0+{\rho }_{l}g(H-y)}{P_0+{\rho }_{l}gH}{)}^{2/5}$
• C. $T_0(\frac{P_0+{\rho }_{l}gH}{P_0+{\rho }_{l}gy}{)}^{3/5}$
• D. $T_0(\frac{P_0+{\rho }_{l}g(H-y)}{P_0+{\rho }_{l}gH}{)}^{3/5}$

Answer 1

The correct option is B $T_0(\frac{P_0+{\rho }_{l}g(H-y)}{P_0+{\rho }_{l}gH}{)}^{2/5}$

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 $\frac{1-\gamma }{\gamma }=\frac{1-\frac{5}{3}}{\frac{5}{3}}=-\frac{2}{5}$
$T_2=T_1(\frac{P_1}{P_2}{)}^{-\frac{2}{5}}=T_1(\frac{P_2}{P_1}{)}^{\frac{2}{5}}=T_0(\frac{P_0+{\rho }_{l}g(H-y)}{P_0+{\rho }_{l}gH}{)}^{\frac{2}{5}}$