Question

If 68.0 cm 3 of oxygen gas is collected over water at 45 °C and 765 mm Hg, what is the volume of dry oxygen gas at STP? The vapor pressure of water at 45 °C is 71.9mm Hg.

Answer 1

Answer : The volume of dry oxygen gas at STP is, $58.8cm^3$

Explanation :

First we have to calculate the pressure of oxygen gas.

Pressure of oxygen gas = Total pressure - Pressure of water

Pressure of oxygen gas = 765 mmHg - 71.9 mmHg

Pressure of oxygen gas = 693.07 mmHg

Now we have yo calculate the volume of dry oxygen gas at STP.

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

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

where,

$P_1$ = initial pressure of gas = 765 mmHg

$P_2$ = final pressure of gas = 760 mmHg  (atSTP)

$V_1$ = initial volume of gas = $68.0cm^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $45^oC=273+45=318K$

$T_2$ = final temperature of gas = $0^oC=273+0=273K$   (at STP)

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

$\frac{765mmHg\times 68.0cm^3}{318K}=\frac{760mmHg\times V_2}{273K}$

$V_2=58.8cm^3$

Therefore, the volume of dry oxygen gas at STP is, $58.8cm^3$