What's wrong with this gas law problem?
Let's derive the appropriate realtionship... 

$$PV = nRT\tag{1}$$

so 

$$\dfrac{PV}{T} = nR\tag{2}$$

but $nR$ is constant for the given sample of gas, thus

$$\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\tag{3}$$

For the sample the volume is 137 ml, the dry gas has a pressure of 753-22=731 mm Hg, and the temperature is 22 °C = 295 °K. 

Now here is a bit of a wrinkle. What is STP? Until 1982 IUPAC defined STP as 760 mm of mercury at 0 °C = 273 °K. After 1982 IUPAC has defined STP as  750.06 mm of mercury at 0 °C = 273 °K.

For old IUPAC STP

$$V_2 = \dfrac{P_1V_1}{T_1}\times\dfrac{T_2}{P_2} = \dfrac{731\times137}{295}\times\dfrac{273}{760} = 121.95 = 122\text{ ml}$$

For current IUPAC STP

$$V_2 = \dfrac{P_1V_1}{T_1}\times\dfrac{T_2}{P_2} = \dfrac{731\times137}{295}\times\dfrac{273}{750} = 123.57 = 124 \text{ ml}$$

In looking at some possible variations, it seems the prof is using the wrong initial pressure, and the old IUPAC definition.

$$V_2 = \dfrac{P_1V_1}{T_1}\times\dfrac{T_2}{P_2} = \dfrac{753\times137}{295}\times\dfrac{273}{760} = 125.62 = 126\text{ ml}$$

NOTE

When collecting a gas over water you must subtract the partial pressure of water at the collection temperature from the atmospheric pressure, but you don't adjust the volume too. 

Also this assumes that the atmospheric pressure given, 753 mm Hg, is for dry air. 
If you were measuring the pressure with a laboratory barometer then you'd have to adjust the measured laboratory pressure for the relative humidity (partial pressure of water).