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,
![\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BT_1%7D%3D%5Cfrac%7BP_2V_2%7D%7BT_2%7D)
where,
= initial pressure of gas = 636 mmHg
= final pressure of gas = ?
= initial volume of gas = 23 mL = 0.023 L
= final volume of gas at STP = 22.4 L
= initial temperature of gas = ![23^oC=273+23=296K](https://tex.z-dn.net/?f=23%5EoC%3D273%2B23%3D296K)
= final temperature of gas at STP = ![0^oC=273+0=273K](https://tex.z-dn.net/?f=0%5EoC%3D273%2B0%3D273K)
Now put all the given values in the above equation, we get:
![\frac{636mmHg\times 0.023L}{296K}=\frac{P_2\times 22.4L}{273K}](https://tex.z-dn.net/?f=%5Cfrac%7B636mmHg%5Ctimes%200.023L%7D%7B296K%7D%3D%5Cfrac%7BP_2%5Ctimes%2022.4L%7D%7B273K%7D)
![P_2=0.602mmHg](https://tex.z-dn.net/?f=P_2%3D0.602mmHg)
Therefore, the pressure of the dry gas at stp is, 0.602 mmHg