H2O is also know as water
Clouds will form due to more water vapor in the air
Answer:
P=atm

Explanation:
The problem give you the Van Der Waals equation:

First we are going to solve for P:


Then you should know all the units of each term of the equation, that is:







where atm=atmosphere, L=litters, K=kelvin
Now, you should replace the units in the equation for each value:

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

Then operate the fraction subtraction:
P=

And finally you can find the answer:
P=atm
Now solving for b:




Replacing units:

Multiplying and dividing units,(please see the second photo below), we have:


