# Question b938e

Jan 9, 2017

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

The molar mass of He ${M}_{H e} = 4 \frac{g}{\text{mol}}$

So number of moles of He in the sample ${n}_{H e} = \frac{w}{M} _ \left(H e\right) = \frac{11.28}{4} = 2.82 m o l$

Now initial volume of the gas ${V}_{1} = 63.2 L$

Initial pressure of the gas ${P}_{1} = 101.325 k P a$

Initial temperature of the gas ${T}_{1} = 273 K$

Final pressure of the gas ${P}_{2} = 98.1 k P a$

Final temperature of the gas ${T}_{2} = \left(32.2 + 273\right) K = 305.2 K$

Final volume of the gas V_2=?

By combined Boyle's and Charl's law equation

$\frac{{P}_{2} {V}_{2}}{T} _ 2 = \frac{{P}_{1} {V}_{1}}{T} _ 1$

$\implies {V}_{2} = \frac{{P}_{1} {V}_{1} {T}_{2}}{{P}_{2} {T}_{1}}$

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

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

$= \overline{{V}_{2}} = {V}_{2} / {n}_{H e} = \frac{72.98}{2.82} \approx 25.9 L$