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`.

Options

A. `10.6xx10^(-6)m^(3)`
B. `5.3xx10^(-6)m^(3)`
C. `2.8xx10^(-6)m^(3)`
D. `15.6xx10^(-6)m^(3)`

Answer

B

Explanation

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)`