pdflatex in a python subprocess on mac

I'm trying to run pdflatex on a .tex file from a Python 2.4.4. subprocess (on a mac):

import subprocess
subprocess.Popen(["pdflatex", "fullpathtotexfile"], shell=True)

which effectively does nothing. However, I can run "pdflatex fullpathtotexfile" in the terminal without issues, generating a pdf. What am I missing?

[EDIT] As suggested in one of the answers, I tried:

return_value = subprocess.call(['pdflatex', '/Users/Benjamin/Desktop/directory/ON.tex'], shell =False)

which fails with:

Traceback (most recent call last):
  File "/Users/Benjamin/Desktop/directory/generate_tex_files_v3.py", line 285, in -toplevel-
    return_value = subprocess.call(['pdflatex', '/Users/Benjamin/Desktop/directory/ON.tex'], shell =False)
  File "/Library/Frameworks/Python.framework/Versions/2.4//lib/python2.4/subprocess.py", line 413, in call
    return Popen(*args, **kwargs).wait()
  File "/Library/Frameworks/Python.framework/Versions/2.4//lib/python2.4/subprocess.py", line 543, in __init__
    errread, errwrite)
  File "/Library/Frameworks/Python.framework/Versions/2.4//lib/python2.4/subprocess.py", line 975, in _execute_child
    raise child_exception
OSError: [Errno 2] No such file or directory

The file does exist and I am able to run pdflatex /Users/Benjamin/Desktop/directory/ON.tex in the Terminal. Note that pdflatex does throw a good number of warnings... but that shouldn't matter, and this also gives the same error:

return_value = subprocess.call(['pdflatex', '-interaction=batchmode', '/Users/Benjamin/Desktop/directory/ON.tex'], shell =False)

Use the convenience function, subprocess.call

You don't need to use Popen here, call should suffice.

For example:

>>> import subprocess
>>> return_value = subprocess.call(['pdflatex', 'textfile'], shell=False) # shell should be set to False

If the call was successful, return_value will be set to 0, or else 1.

Usage of Popen is typically for cases when you want the store the output. For example, you want to check for the kernel release using the command uname and store it in some variable:

>>> process = subprocess.Popen(['uname', '-r'], shell=False, stdout=subprocess.PIPE)
>>> output = process.communicate()[0]
>>> output
'2.6.35-22-generic\n'

Again, never set shell=True.

You might want either:

output = Popen(["pdflatex", "fullpathtotexfile"], stdout=PIPE).communicate()[0]
print output

or

p = subprocess.Popen(["pdflatex" + " fullpathtotexfile"], shell=True)
sts = os.waitpid(p.pid, 0)[1]

(Shamelessly ripped from this subprocess doc page section ).