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InterviewSolution
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An air bubble of volume `1.0 cm^(3)` rises from the bottom of a lake 40 m deep at a temperature of `12^(@) C`. To what volume does it grow when it reaches the surface, which is at a temperature of `35^(@) C`. ? Given `1 atm = 1.01 xx 10^(5) Pa`.A. `10.6xx10^(-6)m^(3)`B. `5.3xx10^(-6)m^(3)`C. `2.8xx10^(-6)m^(3)`D. `15.6xx10^(-6)m^(3)` |
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Answer» Correct Answer - B Using, `P(1)=rho_(20+pgh` Here, `P_(2)=1.013xx10^(5)atm,h=40m` `rho=10^(3)kgm&(3)`(density of water). `g=9.8ms^(2)` `therefore P_(1)=1.013 xx10^(5)+10^(3)xx9.8xx40=493300 Pa` Now, `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` Here, `T_(1)=(12+273)=285K,T_(2)=(32+273)=328K,V_(1)=1xx10^(-6)m^(3)` `V_(2)` is the volume of the air bubble when it reaches the surface `thereforeV_(2)=(P_(1)V_(1)T_(2))/(T_(1)PP_(2))=((493300xx110^(-6)))/(285xx1.013xx10^(5))xx308` `=5.26xx10^(-6)m^(3)=5.3xx10^(-6)m^(3)` |
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