We will have the following:
First, we are given:
[tex]\begin{gathered} V=75.0mL \\ P=1atm\text{ \lparen constant\rparen} \\ T_i=35C \\ T_f=10C \\ \end{gathered}[/tex]Now, we transform these units to the SI, that is:
[tex]\begin{gathered} V=0.075L \\ P=1atm \\ T_i=308.15K \\ T_f=238.15K \end{gathered}[/tex]Now, using the initial values we determine the quantity of the the gas:
[tex]\begin{gathered} PV=nRT\Rightarrow(1atm)(0.075L)=n(0.082057L\ast atm/K\ast mol)(308.15K) \\ \\ \Rightarrow n=\frac{(1atm)(0.075L)}{(0.082057L\ast atm/K\ast mol)(308.15K)}\Rightarrow n=2.966084068\ast10^{-3}mol \\ \\ \Rightarrow n\approx3.0\ast10^{-3}mol \end{gathered}[/tex]Now that we have the number of moles, then we determine the final volume, that is:
[tex]\begin{gathered} V\approx\frac{(3.0\ast10^{-3}mol)(0.082057L\ast atm/K\ast mol)(238.15K)}{(1atm)}\Rightarrow V\approx0.05862562365L \\ \\ \Rightarrow V\approx0.06L \end{gathered}[/tex]So, the final volume will be approximately 0.06 L.
