Question 2.SP.6: (a) Calculate the density, specific weight, and specific vol......

Table A.5 for oxygen \left(\mathrm{O}_{2}\right) : Molar mass M=32.0, k=1.40

(a) Sec. 2.7: \quad R \approx \frac{R_{0}}{M}=\frac{49,709}{32.0}=1553 \mathrm{ft} \cdot \mathrm{lb} /\left(\mathrm{slug} \cdot{ }^{\circ} \mathrm{R}\right) \quad (as in Table A.5)

From Eq. (2.4): \quad \rho=\frac{p}{R T}=\frac{15 \times 144 \mathrm{lb} / \mathrm{ft}^{2}}{\left[1553 \mathrm{ft} \cdot \mathrm{lb} /\left(\mathrm{slug} \cdot{ }^{\circ} \mathrm{R}\right)\right]\left[(460+100)^{\circ} \mathrm{R}\right]}

=0.00248 \mathrm{slug} / \mathrm{ft}^{3} \quad

With g=32.2 \mathrm{ft} / \mathrm{sec}^{2}, \gamma=\rho g=0.00248(32.2)=0.0800 \mathrm{lb} / \mathrm{ft}^{3} \quad

Eq. (2.2): \quad v=\frac{1}{\rho}=\frac{1}{0.00248}=403 \mathrm{ft}^{3} / \mathrm{slug}

(b) Isentropic compression: v_{2}=40 \% v_{1}=0.4(403)=161.1 \mathrm{ft}^{3} / \mathrm{slug}

\rho_{2}=1 / v_{2}=0.00621 \mathrm{slug} / \mathrm{ft}^{3}

Eq. (2.6) with n=k : p v^{k}=(15 \times 144)(403)^{1.4}=\left(p_{2} \times 144\right)(161.1)^{1.4}

\begin{aligned}& p_{2}=54.1 \text { psia } \end{aligned}

From Eq. (2.4): \quad p_{2}=54.1 \times 144 psia =\rho R T=0.00621(1553)\left(460+T_{2}\right)

T_{2}=348^{\circ} \mathrm{F} \quad

(c) Isothermal compression: T_{2}=T_{1}=100^{\circ} \mathrm{F} \quad

p v= constant: (15 \times 144)(403)=\left(p_{2} \times 144\right)(0.4 \times 403)

p_{2}=37.5 \mathrm{psia} \quad

 v=\frac{1}{\rho}          (2.2)

\frac{p}{\rho}=p v=R T       (2.4)

p v^n=p_1 v_1^n=\text { constant }      (2.6a)

 \frac{p}{p_1}=\left(\frac{\rho}{\rho_1}\right)^n=\text { constant }      (2.6b)