A) Given,
P1 = 3 atm
T1 = 17 degree Celsius = 290 K
V1 = 10 L
P2 = ?
T2 = 47 degree Celsius = 320 K
V2 = 10 L
We know, at constant volume:
P1/P2=T1/T2
P2 = P1T2/T1 =(3x 320)/290 =3.31 atm
B) Now, pressure to be restored = 3 atm
So , decrease in pressure = 3.31 - 3 = 0.31 atm
Let v be the volume of air that is let out
Therefore, at constant temperature,
P1V1 = P2V2
0.31 x 10 = 1 x v
v = 3.1 L