The complete question is this: This figure (Figure 1) shows a container that is sealed at the top by a movable piston. Inside the container is an ideal gas at 1.00 atm, 20.0 ∘C, and 1.00 L. This information will apply to the first three parts of this problem.
A) What will the pressure inside the container become if the piston is moved to the 1.20 L mark while the temperature of the gas is kept constant?
Explanation:
It is given that,
[tex]P_{1}[/tex] = 1 atm, [tex]V_{1}[/tex] = 1 L
[tex]P_{2}[/tex] = ? , [tex]V_{2}[/tex] = 1.20 L
As the temperature is constant. Hence, find the value of [tex]P_{2}[/tex] as follows.
[tex]P_{1} \times V_{1}[/tex] = [tex]P_{2} \times V_{2}[/tex]
[tex]1 atm \times 1 L[/tex] = [tex]P_{2} \times 1.20 L[/tex]
[tex]P_{2}[/tex] = 0.833 atm
Thus, we can conclude that the pressure inside the container is 0.833 atm.
