Question
If $Z$ is a compressibility factor, van der Waals equation at low pressure can be written as:[Main 2014](a) $Z=1+\frac{R T}{P b}$(b) $Z=1-\frac{a}{V R T}$(c) $Z=1-\frac{P b}{R T}$(d) $Z=1+\frac{P b}{R T}$
Step 1
Step 1: First, we start with the van der Waals equation for one mole of gas, which is given by: \[P + \frac{a}{V^2} \cdot (V - b) = RT\] Show more…
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If $Z$ is a compressibility factor, van der Waals equation at low pressure can be written as: |2014] (a) $\mathrm{Z}=1-\frac{\mathrm{Pb}}{\mathrm{RT}}$ (b) $\mathrm{Z}=1+\frac{\mathrm{Pb}}{\mathrm{RT}}$ (c) $\mathrm{Z}=1+\frac{\mathrm{RT}}{\mathrm{Pb}}$ (d) $Z=1+\frac{a}{V R T}$
At low pressures, van der Waals' equation is written as $\left(P+\frac{a}{V^{2}}\right) V=R T .$ The compressibility factor is then equal to: (a) $\left(1-\frac{a}{R T V}\right)$ (b) $\left(1-\frac{R T V}{a}\right)$ (c) $\left(1+\frac{a}{R T V}\right)$ (d) $\left(1+\frac{R T V}{a}\right)$
At low pressure, the van der Waals equation become : (a) $P V_{m}=R T$ (b) $P\left(V_{m}-b\right)=R T$ (c) $\left(P+\frac{a}{V_{m}^{2}}\right) V_{m}=R T$ (d) $P=\frac{R T}{V_{m}}+\frac{a}{V_{m}^{2}}$
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