A diver at a depth of 10 m exhales a bubble of air of volume 24.63 mL. The bubble catches an organism which survives on the exhaled air trapped in the bubble. Find out what will be the volume [ x in mL of the bubble when it reaches the surface after 10 min. The organism just inhales the air at the rate of 0.05 millimoles per minute and exhales nothing.Also find out the average rate [y in m mol/min] at which organism should inhale so that volume of bubble remains constant at the depth and the surface. Hence, find the value of xy excluding decimal places.[Given:P atm=1atm; d(H2O)=1""g/cm3;g=1000 cm/s2; T(H2O)=300""K(throughout)]

Question

A diver at a depth of 10 m exhales a bubble of air of volume 24.63 mL. The bubble catches an organism which survives on the exhaled air trapped in the bubble. Find out what will be the volume [ x in mL of the bubble when it reaches the surface after 10 min. The organism just inhales the air at the rate of 0.05 millimoles per minute and exhales nothing.Also find out the average rate [y in m mol/min] at which organism should inhale so that volume of bubble remains constant at the depth and the surface. Hence, find the value of xy excluding decimal places.[Given:P atm=1atm; d(H2O)=1""g/cm3;g=1000 cm/s2; T(H2O)=300""K(throughout)]

Accepted Answer

Pair inside the bubble =1+100×1×1000105=2atmnair inside the bubble  =2×24.63×10−30.0821×300=2×10−3 moles After 10 min, nair inside =2−0.05×10=1.5m molP1V1n1=P2V2n22×24.632=1×V21.5⇒V2=36.945mLNow, P1n1=P2n2⇒    22=12−10r⇒    r=0.1mmol/min.⇒    yx=36.9450.1=369.45

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