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Answer 1 · 2017-01-09T10:40:37

Here the mass of He gas $w=11.28g$

The molar mass of He $M_(He)=4g/"mol"$

So number of moles of He in the sample $n_(He)=w/M_(He)=11.28/4=2.82mol$

Now initial volume of the gas $V_1=63.2L$

Initial pressure of the gas $P_1=101.325kPa$

Initial temperature of the gas $T_1=273K$

Final pressure of the gas $P_2=98.1kPa$

Final temperature of the gas $T_2=(32.2+273)K=305.2K$

Final volume of the gas $V_2=?$

By combined Boyle's and Charl's law equation

$(P_2V_2)/T_2=(P_1V_1)/T_1$

$=>V_2=(P_1V_1T_2)/(P_2T_1)$

$=>V_2=(101.325xx63.2xx305.2)/(98.1xx2 73L)~~72.98L$

So molar volume at $32.2^@C$

$=bar(V_2)=V_2/n_(He)=72.98/2.82~~25.9L$