Chemistry - Ideal Gas Question with Solution | TestHub
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 l-atm/K-mol, g = 10 ms-2)
ab = volume (in ml) of bubble when it reaches the surface after 10 min
cd = average rate (in 10-5 mole/min) at which the organism should inhale air so that volume of bubble remains the same at the depth and at the surface.
The value of abcd is
Answer:
Solution:
Pressure at 10 m depth: P = Patm + rho g h = 1 atm + (1013 × 10 × 10) Pa = 1 atm + 1.013 × 105 Pa = 2 atm. Initial moles in bubble: n0 = PV/RT = (2 × 0.024)/(0.08 × 300) = 2.0 × 10-3 mol = 2.0 mmol. In 10 min the organism consumes 0.05 × 10 = 0.5 mmol, leaving 1.5 mmol. Volume at the surface (P = 1 atm): V = nRT/P = (1.5 × 10-3)(0.08)(300)/1 = 0.036 L = 36 ml, so ab = 36. For the bubble to have the same 24 ml at the surface, the moles at the surface must be n = (1 × 0.024)/(0.08 × 300) = 1.0 mmol, i.e. the organism must consume 1.0 mmol in 10 min = 0.1 mmol/min = 10 × 10-5 mol/min, so cd = 10. Hence abcd = 3610.
