Answer
$T_{m}=16.2^{∘}C$
$P_{m}=138.9\ kPa$
Work Step by Step
The mole number of each gas is
$$
\begin{aligned}
& N_{\mathrm{Ne}}=\left(\frac{P_1 V_1}{R_u T_1}\right)_{\mathrm{Ne}}=\frac{(100 \mathrm{kPa})\left(0.45 \mathrm{~m}^3\right)}{\left(8.314 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kmol} \cdot \mathrm{K}\right)(293 \mathrm{~K})}=0.0185 \mathrm{kmol} \\
& N_{\mathrm{Ar}}=\left(\frac{P_1 V_1}{R_u T_1}\right)_{\mathrm{Ar}}=\frac{(200 \mathrm{kPa})\left(0.45 \mathrm{~m}^3\right)}{\left(8.314 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kmol} \cdot \mathrm{K}\right)(323 \mathrm{~K})}=0.0335 \mathrm{kmol}
\end{aligned}
$$ Thus,
$$ N_m=N_{\mathrm{Ne}}+N_{\mathrm{Ar}}=0.0185 \mathrm{kmol}+0.0335 \mathrm{kmol}=0.0520 \mathrm{kmol}
$$ (a) We take both gases as the system. No work or mass crosses the system boundary, therefore this is a closed system with $W=0$. Then the conservation of energy equation for this closed system reduces to
$$
\begin{aligned}
E_{\text {in }}-E_{\text {out }} & =\Delta E_{\text {system }} \\
-Q_{\text {out }} & =\Delta U=\Delta U_{\mathrm{Ne}}+\Delta U_{\mathrm{Ar}} \longrightarrow-Q_{\text {out }}=\left[m c_v\left(T_m-T_1\right)\right]_{\mathrm{Ne}}+\left[m c_v\left(T_m-T_1\right)\right]_{\mathrm{Ar}}
\end{aligned}
$$ Using $c_v$ values at room temperature and noting that $m=N M$, the final temperature of the mixture is determined to be $$
\begin{aligned}
-15 \mathrm{~kJ} & =(0.0185 \times 20.18 \mathrm{~kg})\left(0.6179 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(T_m-20^{\circ} \mathrm{C}\right) \\
& +(0.0335 \times 39.95 \mathrm{~kg})\left(0.3122 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(T_m-50^{\circ} \mathrm{C}\right) \\
T_m & =16.2^{\circ} \mathrm{C}(289.2 \mathrm{~K})
\end{aligned}
$$ (b) The final pressure in the tank is determined from $$
P_m=\frac{N_m R_{u t} T_m}{V_m}=\frac{(0.052 \mathrm{kmol})\left(8.314 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kmol} \cdot \mathrm{K}\right)(289.2 \mathrm{~K})}{0.9 \mathrm{~m}^3}=\mathbf{1 3 8 . 9} \mathbf{~ k P a}
$$
