This problem deals with only one set of conditions, a situation w here the ideal gas equation is applicable. Three of the four variables in the ideal gas equation (P, n, and T) are given, and the fourth (V) is to be calculated.

P = 0.992 atm        n = 1.52 moles
V = ? L                  T = 65°C = 338 K

Rearranging the ideal gas equation to isolate V on the left side of the equation gives

V = \frac{nRT}{P}

Since the pressure is given in atmospheres and the volume unit is in liters, the appropriate R value is

R = 0.08206 \frac{\textrm{atm} \cdot \textrm{L}}{\textrm{mole} \cdot \textrm{K}}

Substituting the given numerical values into the equation and cancelling units gives

V = \frac{(1.52  \cancel{\textrm{moles}}) \Bigl(0.08206  \frac{ \cancel{\textrm{atm}} \cdot \textrm{L}}{\cancel{\textrm{mole}} \cdot \cancel{\textrm{K}}} \Bigr) (338  \cancel{\textrm{K}})}{0.992  \cancel{\textrm{atm}}}

Note how all of the parts of the ideal gas constant unit system cancel except for one, the volume part.
Doing the arithmetic, we get as an answer 42.5 L CO.

V = \frac{1.52  \times 0.08206 \times 338}{0.992} L CO

= 42.499138 L CO        (calculator answer)
= 42.5 L CO                  (correct answer)

Answer Double Check: Is the ideal gas law, in its rearranged form, in a form such that all of the units cancel except for that of volume (liters)? The answer is yes.