Question

At an ocean depth of 10.0m, a diver's lung capacity is 2.40L. The air temperature is 32.0°C and the pressure is 101.30 kPa. What is the volume of the diver's at the same depth, at a temperature of 21.0°C and a pressure of 141.20 kPa?

Answer 1

Answer: The volume of the diver's at the same depth, at a temperature of 21.0°C and a pressure of 141.20 kPa is 1.66 L

Step-by-step explanation:

The combined gas equation is,

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

where,

`P_1` = initial pressure of gas = 101.30 kPa

`P_2` = final pressure of gas = 141.20 kPa

`V_1` = initial volume of gas = 2.40 L

`V_2` = final volume of gas = ?

`T_1` = initial temperature of gas =
`32.0^0C=(32+273)K=305K`

`T_2` = final temperature of gas =
`21.0^0C=(21+273)K=294K`

Now put all the given values in the above equation, we get:

`(101.30* 2.40)/(305)=(141.20* V_2)/(294)`

`V_2=1.66L`

The volume of the diver's at the same depth, at a temperature of 21.0°C and a pressure of 141.20 kPa is 1.66 L