Question

Question #b938e

Answer 1

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`