Answer: -8.51 ft^3/lbm
Explanation:
In this question, to find the change in volume, we will use the ideal gas law.
[tex]\begin{aligned}&n_{1}=1 \text { mole }\\&\mathrm{P}_{1}=20 \mathrm{psia}\\&\mathrm{T}_{1}=70^{\circ} \mathrm{F}\\&T_{1}=(70+460)^{0} \mathrm{R}\\&v_{1}=\frac{\left(0.3704 \frac{psia \cdot_{\text { } ft^{3}}}{l \mathrm{lbm} \cdot R}\right)(70+460)^{0} \mathrm{R}}{20 \mathrm{psi}}\\&v_{1}=9.816 \frac{\mathrm{ft}^{3}}{\mathrm{lbm}}\end{aligned}[/tex]
[tex]\begin{aligned}&\mathrm{n}_{2}=1 \text { mole } \\&\mathrm{P}_{2}=150 \mathrm{psia} \\&\mathrm{T}_{1}=\mathrm{T}_{2}=70^{0} \mathrm{~F} \\&\mathrm{~T}_{1}=\mathrm{T}_{2}=(70+460)^{0} \mathrm{R} \\&v_{2}=\frac{\left(0.3704 \frac{\text { psia} \cdot \mathrm{ft}^{3}}{\mathrm{lbm} \cdot R}\right)(70+460)^{0} \mathrm{R}}{150 \mathrm{psi}} \\&v_{2}=1.309 \frac{\mathrm{ft}{ }^{3}}{\mathrm{lbm}}\end{aligned}[/tex]
[tex]$The change in volume is,$\begin{aligned}&\Delta v=v_{2}-v_{1} \\&\Delta v=1.309-9.816 \\&\Delta v=-8.51 \frac{\mathrm{ft}^{3}}{\mathrm{lbm}}\end{aligned}\text{The negative sign shows that the gas was compressed during this process}[/tex]
