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

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

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:

$\frac{101.30\times 2.40}{305}=\frac{141.20\times V_2}{294}$

$V_2=1.66L$

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