Question
Question: The density of water is \( 1.0\dfrac{g}{{mL}} \) . What is the density of water in \( \dfrac{{pounds...
The density of water is 1.0mLg . What is the density of water in gallonpounds ?
Solution
The question of dimensional analysis. You have to convert mLg into gallonpounds that is mL1g×??=galpounds . The question mark is the conversion factor that you need to find out to convert. Remember that 1 ponds=453.6g and 1 gallon=3785.411 mL . Use these values to convert grams per milliliters to pounds per gallons.
Complete Step By Step Answer:
This is the question of dimensional analysis. The study of the relationship between physical quantities with the help of dimensions and units of measurement is termed dimensional analysis. The basic concept of dimension is that we can add and subtract only those quantities that have the same dimensions.
You need to convert mLg into gallonpounds . Find a conversion factor that can be multiplied with the mLg units to give the value in gallonpounds .
Here mass conversation unit g→pounds and volume conversation unit mL→gal is required. It is known that for mass, 1 ponds=453.6g and for volume 1 gallon=3785.411 mL . For our solving purpose we will write it as: 453.6g1pounds and 1gal3785.411 mL . These two will be our conversion factors that will be multiplied with the mLg unit.
Now that we know the conversion factors to be multiplied, we will apply them and find out the final result.
mL1g×453.6g1pounds×1gal3785.411mL=gal8.34pounds
Therefore the density of water in gallonpounds is gal8.34pounds .
Note:
Dimensional analysis has limitations also. It doesn’t give information about the dimensional constant. It gives no information about whether a physical quantity is a scalar or vector. The formula containing trigonometric function, exponential functions, logarithmic function, etc. cannot be derived.