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Answer 1 · 2017-01-09T10:40:37

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

The molar mass of He $M_{H e} = 4 \frac{g}{mol}$ $M_(He)=4g/"mol"$

So number of moles of He in the sample $n_{H e} = \frac{w}{M_{H e}} = \frac{11.28}{4} = 2.82 m o l$ $n_(He)=w/M_(He)=11.28/4=2.82mol$

Now initial volume of the gas $V_{1} = 63.2 L$ $V_1=63.2L$

Initial pressure of the gas $P_{1} = 101.325 k P a$ $P_1=101.325kPa$

Initial temperature of the gas $T_{1} = 273 K$ $T_1=273K$

Final pressure of the gas $P_{2} = 98.1 k P a$ $P_2=98.1kPa$

Final temperature of the gas $T_{2} = (32.2 + 273) K = 305.2 K$ $T_2=(32.2+273)K=305.2K$

Final volume of the gas $V_{2} = ?$ $V_2=?$

By combined Boyle's and Charl's law equation

$\frac{P_{2} V_{2}}{T_{2}} = \frac{P_{1} V_{1}}{T_{1}}$ $(P_2V_2)/T_2=(P_1V_1)/T_1$

$\Rightarrow V_{2} = \frac{P_{1} V_{1} T_{2}}{P_{2} T_{1}}$ $=>V_2=(P_1V_1T_2)/(P_2T_1)$

$\Rightarrow V_{2} = \frac{101.325 \times 63.2 \times 305.2}{98.1 \times 273 L} \approx 72.98 L$ $=>V_2=(101.325xx63.2xx305.2)/(98.1xx2 73L)~~72.98L$

So molar volume at ${32.2}^{\circ} C$ $32.2^@C$

$= ¯ ¯¯ ¯ V_{2} = \frac{V_{2}}{n_{H e}} = \frac{72.98}{2.82} \approx 25.9 L$ $=bar(V_2)=V_2/n_(He)=72.98/2.82~~25.9L$