A glass tumbler containing 243 cm^(3) of air at 10^(2) kpa (the barometric pressure) at 20^(@) C is turned upside down and immersed in a body of water to a depth of 20.0 m. The air in the glass is commpressed by weight above it. Calculate the volume of air in glass, assuming temperature and barometric pressure are constant .

A glass tumbler containing 243 cm^(3) of air at 10^(2) kpa (the barome

1.	A glass tumbler containing 243 cm^(3) of air at 10^(2) kpa (the barometric pressure) at 20^(@) C is turned upside down and immersed in a body of water to a depth of 20.0 m. The air in the glass is commpressed by weight above it. Calculate the volume of air in glass, assuming temperature and barometric pressure are constant .
Answer» Solution :`P_(f) = P_(1) +` Pressure due to water column `P_("due to water column") = dgh` `= (1 XX 10^(-3))/(10^(-3) xx 10^(-3)) (KG)/(m^(3)) xx 9.8 xx 20.5` `= 200.9 k Pa` `P_(f) = (100 + 200.9) k Pa = 300.9 k Pa` Using Boyle's Law `P_(1)V_(1) = P_(2) V_(2)` `V_(2) = (100 xx243)/(300.9) = 80.75 ML` `= (80.75)/(20) = 4`