Answer
A) 722.6 mL
Procedure
To solve this problem we can use the ideal gas Law given the temperature and pressure conditions.
Data
R= 0.08206L⋅atm⋅K⁻¹⋅mol⁻¹
P= 1.20 atm
T=112 °C =385.15 °K
V=0.735 L
Formula
PV=nRT
Solving for n to get the moles we will have
[tex]n=\frac{PV}{RT}=\frac{1.20\text{ atm }0.735\text{ L }\degree\text{K.mol}}{0.08206\text{ L.atm }385.15\text{ }\degree\text{K}}=0.02791\text{ mol}[/tex]Once we have calculated the moles we solve again with the new temperature and pressure conditions, remember to use the units of R constant. Pressure in atm and temperature in Kelvin for this case.
Data
n=0.02791 mol
R= 0.08206L⋅atm⋅K⁻¹⋅mol⁻¹
P= 660 mmHg = 0.868421 atm
T=274 °K
Solve for V
[tex]V=\frac{nRT}{P}=\frac{0.02791\text{ mol }0.08206\text{ L.atm }274\degree\text{K}}{0.868421\text{ atm mol }\degree\text{K}}=0.7226\text{ L}=722.6\text{ mL}[/tex]