A diver at a depth of 10 m exhales some air by which a bubble of air of volume 24.0 ml is formed. Th — States of Matter and Gaseous State Chemistry Question
Question
A diver at a depth of 10 m exhales some air by which a bubble of air of volume 24.0 ml is formed. The bubble catches an organism which survives on the exhaled air trapped in the bubble. The organism just inhales the air at the rate of 0.05 millimoles/min and exhales nothing. The atmospheric pressure is 1 atm and the temperature of water is throughout 300 K. Density of water is 1.013 g/ml. ($R = 0.08 \text{ l-atm/K-mol}$, $g = 10 \text{ ms}^{-2}$)<br>$ab = \text{volume (in ml) of bubble when it reaches the surface after 10 min}$.<br>$cd = \text{average rate (in } 10^{-5} \text{ mole/min) at which the organism should inhale air so that volume of bubble remains the same at the depth and at the surface}$.<br>The value of $abcd$ is
💡 Solution & Explanation
Pressure at 10m depth $P_1 = P_{atm} + \rho g h = 101325 + 1013 \times 10 \times 10 \approx 2.0\text{ atm}$. Initial volume $V_1 = 24.0\text{ ml}$. Moles $n_1 = \frac{2.0 \times 0.024}{0.08 \times 300} = 2.0\text{ mmol}$. Organism consumes $0.05 \times 10 = 0.5\text{ mmol}$, leaving $n_2 = 1.5\text{ mmol}$. At surface, $P_2 = 1.0\text{ atm}$. $V_2 = \frac{1.5 \times 10^{-3} \times 0.08 \times 300}{1.0} = 36.0\text{ ml}$. Thus $ab = 36$. For volume to remain 24.0 ml at surface, $n_2$ must be $1.0\text{ mmol}$. Total consumed $= 2.0 - 1.0 = 1.0\text{ mmol}$ in 10 min. Rate $= 0.1\text{ mmol/min} = 10 \times 10^{-5} \text{ mol/min}$. Thus $cd = 10$. The value of $abcd$ is 3610. Therefore, correct answer is 3610.