According to the explanation given in the previous session about Charles's law, now we have different temperatures:
We have in situation 1:
V1 = 0.510 L
T1 = 292 K
V2 = ?
T2 = 422 K
In situation 2:
V1 = 0.510 L
T1 = 292 K
V2 = ?
T2 = 260 K
In situation 3:
V1 = 0.510 L
T1 = 292 K
V2 = ?
T2 = 574 K
Now solving each situation:
1.
0.510/292 = V2/422
0.00175 = V2/422
V2 = 0.738 Liters
2.
0.510/292 = V2/260
0.00175 = V2/260
V2 = 0.455 Liters
3.
0.510/292 = V2/574
0.00175 = V2/574
V2 = 1.00 Liters
