Because there is a volume of 50 L, maximum mass of 975 g, and propane has a molecular weight MW=44.09 g/mol, the molar volume of propane in this process can be calculated.
\underline{\mathrm{V}}=(50 \mathrm{~L})\left(\frac{\mathrm{m}^3}{1000 \mathrm{~L}}\right)\left(\frac{100^3 \mathrm{~cm}^3}{\mathrm{~m}^3}\right)\left(\frac{1}{975 \mathrm{~g}}\right)\left(\frac{44.09 \mathrm{~g}}{\mathrm{~mol}}\right)=2261 \frac{\mathrm{cm}^3}{\mathrm{~mol}}
A) The ideal gas law can be rearranged to solve for the maximum pressure of the tank under these specifications.
\mathrm{P} \underline{\mathrm{V}}=\mathrm{RT}
\mathrm{P}=\frac{\mathrm{RT}}{\underline{\mathrm{V}}}
Solve for pressure using the maximum temperature of 425 K and the previously calculated molar volume
P=\frac{\left[(425 \mathrm{~K})\left(83.14 \frac{\mathrm{cm}^3 \mathrm{bar}}{\mathrm{m} \mathrm{ol} \,\mathrm{K}}\right)\right]}{2261 \frac{\mathrm{cm}^3}{\mathrm{~mol}}}
\bf P=15.63 \,bar
Violates the policy of not exceeding half of the burst pressure of 30 bar
B) from Appendix C:
for propane \rightarrow \mathrm{T}_{\mathrm{c}}=369.83 \mathrm{~K} ; \mathrm{P}_{\mathrm{c}}=42.48~ \mathrm{bar} ; \omega=0.152
Using \rm T_c and \rm P_c, calculate the reduced temperature and pressure at 425 K and 30 bar
\begin{aligned}& \mathrm{T}_{\mathrm{r}}=\frac{\mathrm{T}}{\mathrm{T}_{\mathrm{c}}} \\& \mathrm{T}_{\mathrm{r}}=\frac{425 \mathrm{~K}}{369.83 \mathrm{~K}}=1.15 \\& \mathrm{P}_{\mathrm{r}}=\frac{\mathrm{P}}{\mathrm{P}_{\mathrm{c}}}\end{aligned}
\mathrm{P}_{\mathrm{r}}=\frac{30 \,\mathrm{bar}}{42.48 \,\mathrm{bar}}=0.71
From the Lee Kesler charts, find z^0 and z^1 using the reduced temperature and pressure calculated in the previous step
For propane at \mathrm{T}_{\mathrm{r}}=1.15 and \mathrm{P}_{\mathrm{r}}=0.71 \rightarrow \mathrm{z}^0=0.8 ; \mathrm{z}^1=0.025
Calculate z
\begin{aligned}& \mathrm{z}=\mathrm{z}^0+\omega \mathrm{z}^1 \\& \mathrm{z}=0.8+(0.152)(0.025)=0.8038\end{aligned}
Solve for molar volume using z
\begin{aligned}& \mathrm{PV}=\mathrm{zRT} \\& \mathrm{P}=\frac{\mathrm{zRT}}{\underline{\mathrm{V}}}\end{aligned}
\begin{gathered}P=\frac{(0.8038)\left(83.14 \frac{\mathrm{cm}^3 \mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}\right)(425 \mathrm{~K})}{2261 \frac{\mathrm{cm}^3}{\mathrm{~mol}}} \\\mathrm{P}=12.56 \,\mathrm{bar}\end{gathered}
Not in violation because the pressure is less than half the busting pressure
C) The Virial Equation is expressed in terms of z:
\begin{aligned}& \mathrm{z}=1+\left(\frac{\mathrm{BP}_{\mathrm{c}}}{\mathrm{RT}_{\mathrm{c}}}\right)\left(\frac{\mathrm{P}_{\mathrm{r}}}{\mathrm{T}_{\mathrm{r}}}\right) \\& \left(\frac{\mathrm{BP}_{\mathrm{c}}}{\mathrm{RT}_{\mathrm{c}}}\right)=\mathrm{B}^0+\omega \mathrm{B}^1\end{aligned}
Use the reduced temperature found in part b
\begin{aligned}& \mathrm{B}^0=0.083-\frac{0.422}{\mathrm{~T}_{\mathrm{r}}^{1.6}} \\& \mathrm{~B}^0=0.083-\frac{0.422}{1.15^{1.6}}=-0.254 \\& \mathrm{~B}^1=0.139-\frac{0.172}{\mathrm{~T}_{\mathrm{r}}^{4.2}} \\& \mathrm{~B}^1=0.139-\frac{0.172}{1.15^{4.2}}=0.0434 \\& \left(\frac{\mathrm{BP}_{\mathrm{c}}}{\mathrm{RT}_{\mathrm{c}}}\right)=-0.254+(0.152)(0.0434)=-0.247\end{aligned}
Rewrite the Virial Equation in terms of pressure
\mathrm{z}=\frac{\mathrm{P}\underline{\mathrm{V}}}{\mathrm{RT}}=1+\left(\frac{\mathrm{BP}_{\mathrm{c}}}{\mathrm{RT}_{\mathrm{c}}}\right)\left(\frac{\mathrm{P}}{\mathrm{P}_{\mathrm{c}} \mathrm{T}_{\mathrm{c}}}\right)
Solve for pressure using all known values.
\frac{\left(2261 \frac{\mathrm{cm}^3}{\mathrm{~mol}}\right) \mathrm{P}}{(425 \mathrm{~K})\left(83.14 \frac{\mathrm{cm}^3 \mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}\right)}=1-0.247\left(\frac{\mathrm{P}}{(42.48 \,\mathrm{bar})(369.83 \mathrm{~K})}\right)
P = 14.48 < 30 bar
Not in violation because the pressure in the tank does not exceed half the bursting pressure