Answer:
[tex]V_2=1.34L[/tex]
Explanation:
Hello there!
In this case, since this problem is describing how pressure changes as a function of volume and vice versa, it is possible recall the Boyle's law as shown below:
[tex]P_1V_1=P_2V_2[/tex]
Whereas we are asked to compute the volume when the change is pressure is performed (V₂); thus, we proceed as follows:
[tex]V_2=\frac{P_1V_1}{P_2}\\\\V_2=\frac{1.55L*105.2kPa}{122.0kPa}\\\\V_2=1.34L[/tex]
Best regards!