Question

question 8
a 6.75 - l flask contains 6.49 mol he at 310°c. calculate the pressure of this sample of he from the van der waals equation.
type numbers in
10 points
atm
note: for he, a = 0.0342 l² atm/mol² and b = 0.0237 l/mol
round your answer to 3 significant figures.

Explanation:

Step1: Convert temperature to Kelvin

$$T = 310 + 273.15=583.15\ K$$

Step2: Substitute values into van der Waals equation

The van der Waals equation is \((P+\frac{n^{2}a}{V^{2}})(V - nb)=nRT\).
We need to solve for \(P\). First, expand the left - hand side:
\(P(V - nb)+\frac{n^{2}a}{V^{2}}(V - nb)=nRT\)
\(P(V - nb)=nRT-\frac{n^{2}a}{V^{2}}(V - nb)\)
\(P=\frac{nRT}{V - nb}-\frac{n^{2}a}{V^{2}}\)

Given \(n = 6.49\ mol\), \(V=6.75\ L\), \(a = 0.0342\ L^{2}\ atm/mol^{2}\), \(b = 0.0237\ L/mol\), \(R = 0.0821\ L\cdot atm/mol\cdot K\), \(T = 583.15\ K\)

First, calculate \(nb\): \(nb=6.49\times0.0237 = 0.154813\ L\)

\(V-nb=6.75 - 0.154813=6.595187\ L\)

\(nRT=6.49\times0.0821\times583.15\)
\(nRT=6.49\times47.876615\)
\(nRT = 310.7292\)

\(\frac{n^{2}a}{V^{2}}=\frac{6.49^{2}\times0.0342}{6.75^{2}}\)
\(n^{2}=6.49^{2}=42.1201\)
\(\frac{n^{2}a}{V^{2}}=\frac{42.1201\times0.0342}{45.5625}\)
\(\frac{n^{2}a}{V^{2}}=\frac{1.440507}{45.5625}\approx0.0316\)

\(P=\frac{310.7292}{6.595187}- 0.0316\)
\(P = 47.115-0.0316\)
\(P\approx47.1\ atm\)

Answer:

\(47.1\)