Volume of 0.98 mol of oxygen at S.T.P.= 0.98×22.4 l=21.952 l
Now
Let P1,V1,T1 be the quantities at S.T.P.
V1=21.952P1=1 atmT1=273 KP2=2 atmT2=275 KAlso,T1P1V1=T2P2V2
V2=P2T1P1V1T2=2×2731×21.952×275=11.056 l
Volume occupied by oxygen will be 11.056 l
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