Chemistry - Ideal Gas Question with Solution | TestHub
A diver at a depth of 10 m exhales air forming a bubble of volume 36.0 ml. An organism trapped with the bubble inhales the air at the rate of 0.1 millimoles/min and exhales nothing. The atmospheric pressure is 1 atm and the temperature of the 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 the 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 the volume of the bubble remains the same at the depth and at the surface.
The value of abcd is
Answer:
Solution:
Pressure at 10 m: P = 1 atm + rho g h = 1 atm + (1013 × 10 × 10) Pa = 2 atm. Initial moles: n₀ = PV/RT = (2 × 0.036)/(0.08 × 300) = 3.0 mmol. In 10 min the organism consumes 0.1 × 10 = 1.0 mmol, leaving 2.0 mmol. At the surface (1 atm): V = nRT/P = (2.0 × 10⁻³)(0.08)(300) = 0.048 L = 48 ml, so ab = 48. For the bubble to stay at 36 ml at the surface, the moles must be n = (1 × 0.036)/(0.08 × 300) = 1.5 mmol, i.e. it must consume 1.5 mmol in 10 min = 0.15 mmol/min = 15 × 10⁻⁵ mol/min, so cd = 15. Hence abcd = 4815.
