[Python] OSError [Errno 22] invalid argument when use open() in Python

def choose_option(self):
if self.option_picker.currentRow() == 0:
description = open(":/description_files/program_description.txt","r")
self.information_shower.setText(description.read())
elif self.option_picker.currentRow() == 1:
requirements = open(":/description_files/requirements_for_client_data.txt", "r")
self.information_shower.setText(requirements.read())
elif self.option_picker.currentRow() == 2:
menus = open(":/description_files/menus.txt", "r")
self.information_shower.setText(menus.read())


I am using resource files and something is going wrong when i am using it as argument in open function, but when i am using it for loading of pictures and icons everything is fine.

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