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,
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 636 mmHg
[tex]P_2[/tex] = final pressure of gas = ?
[tex]V_1[/tex] = initial volume of gas = 23 mL = 0.023 L
[tex]V_2[/tex] = final volume of gas at STP = 22.4 L
[tex]T_1[/tex] = initial temperature of gas = [tex]23^oC=273+23=296K[/tex]
[tex]T_2[/tex] = final temperature of gas at STP = [tex]0^oC=273+0=273K[/tex]
Now put all the given values in the above equation, we get:
[tex]\frac{636mmHg\times 0.023L}{296K}=\frac{P_2\times 22.4L}{273K}[/tex]
[tex]P_2=0.602mmHg[/tex]
Therefore, the pressure of the dry gas at stp is, 0.602 mmHg
