Answer:
C.) v=2.20L
Explanation:
According to the Combined Gas Law:
[tex]\frac{p_1v_1}{t_1}=\frac{p_2v_2}{t_2}[/tex]
For this problem, let
[tex]p_1=0.880atm\\v_1=0.600L\\t_1=(46+273)K=319K\\\\p_2=0.205atm\\t_2=(0+273)K=273K[/tex]
So,
[tex]\frac{(0.880atm)(0.600L)}{319K}=\frac{(0.205atm)v_2}{273K}\\0.00166\frac{atm*L}{K}=\frac{(0.205atm)v_2}{273K}\\(0.00166\frac{atm*L}{K})(273K)=(\frac{(0.205atm)v_2}{273K})(273K)\\0.452atm*L=(0.205atm)v_2\\\frac{0.452atm*L}{0.205atm}=\frac{(0.205atm)v_2}{0.205atm}\\2.20L=v_2[/tex]
