Given \( p=130\,\text{kPa}=130\times10^{3}\,\text{Pa}, T=30^{\circ}\text{C}=30+273=303\,\text{K}. \)
The density can be calculated using ideal gas law as,
$$ \boxed{\rho=\frac{p}{R\,T}=\frac{130\times10^{3}}{287\times303}=1.495\,\text{kg/m}^{3}}\ . $$The density in \( \text{lbm/ft}^{3} \) can be calculated using the following conversions \( 1\,\text{kg}=2.2046\,\text{lbm} \), \( 1\,\text{m}=3.2808\,\text{ft} \). Therefore in \( \text{lbm/ft}^{3} \),
$$ \rho=1.495\,\text{kg/m}^{3}=1.495\times2.2046/3.2808^{3}\,\text{lbm/ft}^{3} $$ $$ \boxed{\rho=0.09333\,\text{lbm/ft}^{3}}\ . $$--- Ad ---
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