Question

18 g of helium gas and 25 g of neon gas are mixed in a 3.50 l container at 195 k. what is the total pressure exerted by the gases? 200 atm 16 atm 26 atm 0.038 atm

Explanation:

Step1: Calculate moles of helium

The molar mass of helium (\(He\)) is \(M_{He}=4\space g/mol\). Using the formula \(n = \frac{m}{M}\), for \(m = 18\space g\) of \(He\), \(n_{He}=\frac{18\space g}{4\space g/mol}= 4.5\space mol\)

Step2: Calculate moles of neon

The molar mass of neon (\(Ne\)) is \(M_{Ne}=20.18\space g/mol\). For \(m = 25\space g\) of \(Ne\), \(n_{Ne}=\frac{25\space g}{20.18\space g/mol}\approx1.24\space mol\)

Step3: Calculate total moles

Using \(n_{total}=n_{He}+n_{Ne}\), \(n_{total}=4.5 + 1.24=5.74\space mol\)

Step4: Use ideal gas law \(PV = nRT\)

We know \(V = 3.50\space L\), \(T=195\space K\), \(R = 0.0821\space L\cdot atm/(mol\cdot K)\). Rearranging for \(P\), \(P=\frac{nRT}{V}\)

Substitute \(n = 5.74\space mol\), \(R = 0.0821\space L\cdot atm/(mol\cdot K)\), \(T = 195\space K\), \(V = 3.50\space L\)

\(P=\frac{5.74\times0.0821\times195}{3.50}\)

First calculate \(5.74\times0.0821\times195=(5.74\times0.0821)\times195\approx0.471\times195 = 91.845\)

Then \(P=\frac{91.845}{3.50}\approx26.24\space atm\)

Answer:

26 atm