Answer
$w_{}=139.0 \text{ Btu/lbm}$
Work Step by Step
One lbm of this mixture consists of $0.35$ lbm of nitrogen and $0.65$ lbm of helium or $0.35 \text{lbm}/(28.0 \text{lbm/lbmol}) = 0.0125 \text{ lbmol}$ of nitrogen and $0.65 \text{lbm}/(4.0 \text{lbm/lbmol}) = 0.1625 \text{lbmol}$ of helium. The total mole is $0.0125+0.1625=0.175$ lbmol. The constituent mole fraction are then $$
\begin{aligned}
& y_{\mathrm{N} 2}=\frac{N_{\mathrm{N} 2}}{N_{\text {total }}}=\frac{0.0125 \mathrm{lbmol}}{0.175 \mathrm{lbmol}}=0.07143 \\
& y_{\mathrm{He}}=\frac{N_{\mathrm{He}}}{N_{\text {total }}}=\frac{0.1625 \mathrm{lbmol}}{0.175 \mathrm{lbmol}}=0.9286
\end{aligned}
$$ The effective molecular weight of this mixture is
$$
\begin{aligned}
M & =y_{\mathrm{N} 2} M_{\mathrm{N} 2}+y_{\mathrm{He}} M_{\mathrm{He}} \\
& =(0.07143)(28)+(0.9286)(4) \\
& =5.714 \mathrm{lbm} / \mathrm{bmol}
\end{aligned}
$$ The work done is determined from
$$
\begin{aligned}
w & =\int_1^2 P d \boldsymbol{v}=P_2 v_2-P_1 v_1=R\left(T_2-T_1\right) \\
& =\frac{R_{u t}}{M}\left(T_2-T_1\right)=\frac{1.9858 \mathrm{Btu} / \mathrm{lbmol} \cdot \mathrm{R}}{5.714 \mathrm{lbm} / \mathrm{lbmol}}(500-100) \mathrm{R} \\
& =139.0\ \mathrm{Btu} / \mathrm{lbm}
\end{aligned}
$$
