Does a car consume more fuel when it's raining? Yesterday my wife asked me that question, and I couldn't answer.
Consider a car, in a sunny day, and that is consumes x gallons per mile. Considering that everything is equal, except that it's traveling in a rainy day, but at the same temperature as the sunny day so that air density is the same.
Will the lower friction of the tires make it consume more or less fuel? And the fact that rain drops are falling over and in front of it ?
I answered that it'll consume more fuel, since friction is what makes car move and that the rain will act against it... but I'm not sure ?
 A: I would like to consider just one aspect of this question. I am not sure that, as Chris White put it, "it does not matter how much friction there is so long as your tires are not skidding." The coefficient of rolling friction does depend on the surface, so I guess it is somewhat different in rain. I guess rolling friction losses are somewhat important (http://www.consumerenergycenter.org/transportation/consumer_tips/vehicle_energy_losses.html ), even for driving wheels.
A: I would say yes, I'm in a hurry, but these are the 2 main reasons I can say right now:


*

*Air density is NOT the same, as Chris White noted.

*The water heat capacity and thermal conductivity is different than the air. I can only wildly guess now as there are too many factors involved, but this will likely make it consume more fuel. The engine needs to be hot to deliver it's best and metal is a good conductor. Air not so much, but then water is. So my guess is, even if at the same temperature, the water will make the whole car leak more thermal energy, thus making it consume more energy (fuel). Besides, if you want to keep the car's inside temperature at a different one than the outside one, this will affect it as well.
A: There is an additional loss of energy when driving through puddles on a wet road, because the tire treads have to exert work in order to eject water. One way to look at it is that the keeps trying to glide on top of the water, but is continuously sinking into it to meet the pavement, which is equivalent to driving slightly uphill.

A: Friction is indeed necessary for a car to push off the ground. However, it does not matter how much friction there is so long as your tires are not skidding. You only waste fuel if you put energy into spinning the wheels that is dissipated as heat before ever contributing to the car's motion. Otherwise, Newton's third law holds and the ground pushes you forward as much as you push back. So as long as you're not one of those drivers who keeps hitting the gas in thin puddles and hydroplaning, there shouldn't be much of a difference.
Of course, in order to keep from skidding you may very well have to travel slower, which can itself impact your efficiency (and is probably the dominant effect here - see this related question).
Air resistance is another thing entirely. If there are water droplets in the air, this increases its effective density, so you will spend more fuel pushing it out of the way. "Heavy" rain falls at about $25\ \mathrm{mm}/\mathrm{hr}$ ($1\ \mathrm{in}/\mathrm{hr}$). Elsewhere in that article we learn that large raindrops fall at a terminal velocity of $9\ \mathrm{m}/\mathrm{s}$. In order for these numbers to be consistent, the fractional volume of the air filled with raindrops must be
$$ \chi_V = \frac{25\ \mathrm{mm}/\mathrm{hr}}{9\ \mathrm{m}/\mathrm{s}} = 8\times10^{-7}. $$
As the density of air is $1.2\ \mathrm{kg}/\mathrm{m}^3$, while water is more or less by definition $1000\ \mathrm{kg}/\mathrm{m}^3$, we see the fractional change in air density, assuming constant gas density, is something like
$$ \chi_\rho = \frac{\chi_V(1000\ \mathrm{kg}/\mathrm{m}^3)}{1.2\ \mathrm{kg}/\mathrm{m}^3} = 6\times10^{-4}. $$
So yes, rain will make it harder, by a very small amount, to push through the air.

Addendum: There are many other factors worth considering, many of which have more interesting and complicated physics than the simple stuff I felt like analyzing. Kyle points out in a comment that water can be adhesive, so it can be harder to drive through (this is a less extreme example of driving through tar). Displacing water also costs something, as noted in an answer by kaz. Both of these are probably larger effects than increasing the effective air mass. Frank Presencia Fandos's answer mentions an even larger concern, engine efficiency based on airflow (and readiness to participate in combustion), which is an engineering problem too complicated for a physicist like me. OSE's comment also points out an implicit assumption I made above that should be said explicitly: I'm comparing $100\%$ humidity weather with and without rain. If you want to compare to dry weather, the effect goes the other way, mainly because a molecule of nitrogen is more massive than a molecule of water.
In the end, though, I'm willing to bet the biggest effect is still the fact that most drivers slow down in heavy rain, since fuel efficiency is highly dependent on speed.
A: Driving through standing water takes a lot of energy and wants to slow the car down, so to maintain same speed as on a dry surface, you require more fuel, i.e. your mpg will drop.  I have instant reading mpg on my BMW 335D and have noticed what would other wise be 30mpg on dry surface can drop to 25 or loess on wet surface.  I'm talk surfaces with pooling water, not just wet.
