Question

The compressibility factor for a real gas is expressed by, $z=1+\frac{BP}{RT}$. The value of B at 500 K and 600 bar is 0.0169 L/mol. Find the molar volume of the gas

• A. 0.1 L
• B. $9\times {10}^{-5}L$
• C. $8.62\times {10}^{-2}L/mol$
• D. 1.65 L

Answer 1

The correct option is C $8.62\times {10}^{-2}L/mol$

Given P= 600 bar, T=500 K, B = 0.016 L/mol
By substituting these values in the above equation, we get
$Z=1+\frac{0.0169\times 600}{0.083\times 500}=1.244$
$V_m=\frac{ZRT}{P}=\frac{1.244\times 0.083\times 500}{600}$
or $V_m=8.6\times {10}^{-2}L$