Question

23 ml of oxygen collected over water and has a pressure of 636 mmhg at 23 c.what is the pressure of the dry gas at stp

Answer 1

Answer : The pressure of the dry gas at stp is, 0.602 mmHg

Explanation :

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

The combined gas equation is,

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

where,

`P_1` = initial pressure of gas = 636 mmHg

`P_2` = final pressure of gas = ?

`V_1` = initial volume of gas = 23 mL = 0.023 L

`V_2` = final volume of gas at STP = 22.4 L

`T_1` = initial temperature of gas =
`23^oC=273+23=296K`

`T_2` = final temperature of gas at STP =
`0^oC=273+0=273K`

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

`(636mmHg* 0.023L)/(296K)=(P_2* 22.4L)/(273K)`

`P_2=0.602mmHg`

Therefore, the pressure of the dry gas at stp is, 0.602 mmHg