For 1 mole of an ideal gas, we have PV=RT............(Eq. 1)
where, P = pressure
V= volume
R = universal gas constant
T = Temperature
Let, P1, V1 and T1 be initial pressure, volume and temperature of [tex] O_{2} [/tex] respectively
Let, P2, V2 and T2 be pressure, volume and temperature of [tex] O_{2} [/tex] respectively, after application of pressure.
Using Eq. 1 we get, [tex] \frac{P1V1}{P2V2} = \frac{T1}{T2} [/tex] ....... (Eq. 2)
Now, since that T1=T2 and P2=3(P1)
∴ Eq. 2 becomes P1V1=P2V2
∴ [tex] \frac{V2}{V1} = \frac{P1}{P2} = \frac{1}{3} [/tex]
∴ V2 = [tex] \frac{1}{3} [/tex] V1
Thus, final volume V2 will be 1/3 of initial volume V1