consider a balloon filled with oxygen at 1 atm and at a volume of 10 L at 293K.

Question

consider a balloon filled with oxygen at 1 atm and at a volume of 10 L at 293K. assuming ideal gas, what would be its volume at a depth of 2500 m under the ocean surface at 277K? What is the percentage increase in density of the oxygen?

Explanation / Answer

density = mass/ volume P*V = n*R*T P1 = 1atm P2= P1 + h*density of water* g = 1atm + (2500x1000x9.8/101325)atm =>P2 = 242 atm from ideal gas law equation; P1xV1 = P2xV2 => (1atm)(10L) = (242atm)(V2) => V2 = 0.041L density = mass/ volume.here the mass of gas is constant.so volume xdensity is constant => V1xd1 = V2xd2 =>(d2/d1) = (V1/V2) =>(d2-d1)/d1 = (V1-V2)/V2 =>change in density = (10-0.041)/0.041 = 242.9 times initial density =>% change in density = 24290% the density increases very much in higher depths..hence the huge % change

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