Answer:
[tex]v_2=6,48*10^{-4}L[/tex]
Explanation:
From the question we are told that:
Volume of hydrogen [tex]V_1=30mL=>30*10^{-3}[/tex]
Temperature [tex]T_1=24C=>24+273=>297k[/tex]
Pressure of water [tex]P_1=22.4torr=>\frac{22.4}{760}=0.0295atm[/tex]
Let
STP(Standard Temperature and pressure)
[tex]T_2=273K[/tex]
[tex]P_2=1atm[/tex]
Generally the equation for ideal gas is mathematically given by
[tex]\frac{p_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
Therefore
[tex]v_2=\frac{P_1v_1T_2}{T_1P_2}[/tex]
[tex]v_2=\frac{0.0295*30*10^{-3}*273}{297*1}[/tex]
[tex]v_2=6.48*10^{-4}L[/tex] (at STP(Standard Temperature and pressure))
