To solve this question, we would use the combined gas equation which comprises of Boyle's law, Charles law and pressure law.
The equation is given as
[tex]\begin{gathered} v_1=34mL \\ T_1=21^0C=21+273.15=294.15K \\ P_1=0.953\text{atm} \end{gathered}[/tex]At STP (standard temperature and pressure), the temperature and pressure changes to
[tex]\begin{gathered} p_2=1\text{atm} \\ t_2=273.15K_{} \end{gathered}[/tex]Now, combining the data above, we can find the change in volume (v₂)
[tex]\begin{gathered} \frac{p_1v_1}{t_1}=\frac{p_2v_2}{t_2} \\ v_2=\frac{p_1v_1t_2}{p_2t_1} \\ \end{gathered}[/tex]Let's substitute the values into the equation above and solve for v₂
[tex]\begin{gathered} v_2=\frac{p_1v_1t_2}{p_2t_1} \\ v_2=\frac{0.953\times34\times273.15}{1\times294.15} \\ v_2=30.1mL \end{gathered}[/tex]From the calculations above, the volume at STP is 30.1mL
