Answer:
Boyle's law:
It can be expressed as:
[tex]P_1V_1 =P_2V_2[/tex]
where
[tex]P_1[/tex] and [tex]V_1[/tex] are the initial pressure and initial volume values, and
[tex]P_2[/tex] and [tex]V_2[/tex] are the values of Pressure and volume after changes.
As per the statement:
here,
[tex]P_1 = 1.5 atm[/tex] , [tex]V_1 = 2.0 L[/tex] , [tex]P_2= 3.0 atm[/tex] , [tex]V_1 = A[/tex]
To find the A:
using above formula we have;
[tex]1.5 \cdot 2 = 3A[/tex]
⇒[tex]3 = 3A[/tex]
Divide both sides by 3 we have;
A = 1.0 L
Similarly, find B:
[tex]P_2= B atm[/tex] , [tex]V_2 = 6.0 L[/tex]
Then;
[tex]1.5 \cdot 2.0 = 6B[/tex]
⇒[tex]3 = 6B[/tex]
Divide both sides by 6 we get;
⇒[tex]0.5= B[/tex]
or
B = 0.50 atm
To Find C:
[tex]P_2= C atm[/tex] , [tex]V_2= 5.0 L[/tex]
then;
[tex]1.5 \cdot 2.0 = 5C[/tex]
⇒[tex]3 = 5C[/tex]
Divide both sides by 5 we get;
⇒[tex]0.6= C[/tex]
or
C= 0.60 atm
To Find D:
[tex]P_2= 0.75 atm[/tex] , [tex]V_2= D[/tex]
then;
[tex]1.5 \cdot 2 = 0.75D[/tex]
⇒[tex]3 = 0.75D[/tex]
Divide both sides by 0.75 we have;
D= 4.0 L
Therefore, the values are:
A = 1.0 L
B = 0.50 atm
C = 0.60 atm
D = 4.0 L
