Answer:
Final volumen first process 
Final Pressure second process 
Explanation:
Using the Ideal Gases Law yoy have for pressure:

where:
P is the pressure, in Pa
n is the nuber of moles of gas
R is the universal gas constant: 8,314 J/mol K
T is the temperature in Kelvin
V is the volumen in cubic meters
Given that the amount of material is constant in the process:

In an isobaric process the pressure is constant so:



Replacing : 

Replacing on the ideal gases formula the pressure at this piont is:

For Temperature the ideal gases formula is:

For the second process you have that
So:




Answer:
Planets that are farther from the sun than the earth (all but Mercury and Venus) will exhibit retrograde motion.
If the position of the planet is observed relative to the background stars, the planet will appear to move backward relative to the stars when the earth is moving in an Eastward direction faster than the planet, and the planet appears to move backwards relative to the stars
(The planet will be on the side of the earth that is opposite that of the sun)
"F=Vector Sum Of The Two Forces" Is the answer.