Question

a 10.0 l tank at 10.2°c is filled with 5.85 g of dinitrogen monoxide gas and 6.41 g of dinitrogen difluoride gas. you can assume both gases behave as ideal gases under these conditions.
calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank. be sure your answers have the correct number of significant digits

Explanation:

Step1: Calculate moles of each gas

• For dinitrogen monoxide (\(N_2O\)):

The molar mass of \(N_2O\) is \(M_{N_2O}=(2\times14.01 + 16.00)\space g/mol=44.02\space g/mol\).
Using the formula \(n=\frac{m}{M}\), where \(m = 5.85\space g\) and \(M = 44.02\space g/mol\), we have \(n_{N_2O}=\frac{5.85\space g}{44.02\space g/mol}\approx0.133\space mol\).

• For dinitrogen difluoride (\(N_2F_2\)):

The molar mass of \(N_2F_2\) is \(M_{N_2F_2}=(2\times14.01+2\times19.00)\space g/mol = 66.02\space g/mol\).
Using the formula \(n=\frac{m}{M}\), where \(m = 6.41\space g\) and \(M = 66.02\space g/mol\), we have \(n_{N_2F_2}=\frac{6.41\space g}{66.02\space g/mol}\approx0.0971\space mol\).

Step2: Calculate total moles

Using the formula \(n_{total}=n_{N_2O}+n_{N_2F_2}\), we substitute \(n_{N_2O}=0.133\space mol\) and \(n_{N_2F_2}=0.0971\space mol\).
\(n_{total}=0.133 + 0.0971=0.230\space mol\)

Step3: Calculate mole fractions

• Mole fraction of \(N_2O\):

Using the formula \(x_{N_2O}=\frac{n_{N_2O}}{n_{total}}\), we substitute \(n_{N_2O}=0.133\space mol\) and \(n_{total}=0.230\space mol\).
\(x_{N_2O}=\frac{0.133}{0.230}\approx0.578\)

• Mole fraction of \(N_2F_2\):

Using the formula \(x_{N_2F_2}=\frac{n_{N_2F_2}}{n_{total}}\), we substitute \(n_{N_2F_2}=0.0971\space mol\) and \(n_{total}=0.230\space mol\).
\(x_{N_2F_2}=\frac{0.0971}{0.230}\approx0.422\)

Step4: Convert temperature to Kelvin

Using the formula \(T = t+273.15\), where \(t = 10.2^{\circ}C\).
\(T=(10.2 + 273.15)\space K=283.35\space K\)

Step5: Calculate total pressure

Using the ideal - gas law \(PV=nRT\), where \(V = 10.0\space L\), \(n = n_{total}=0.230\space mol\), \(R = 0.0821\space L\cdot atm/(mol\cdot K)\), and \(T = 283.35\space K\).
\(P_{total}=\frac{n_{total}RT}{V}=\frac{0.230\space mol\times0.0821\space L\cdot atm/(mol\cdot K)\times283.35\space K}{10.0\space L}\)
\(P_{total}=\frac{0.230\times0.0821\times283.35}{10.0}\space atm\approx0.537\space atm\)

Step6: Calculate partial pressures

• Partial pressure of \(N_2O\):

Using the formula \(P_{N_2O}=x_{N_2O}\times P_{total}\), we substitute \(x_{N_2O}=0.578\) and \(P_{total}=0.537\space atm\).
\(P_{N_2O}=0.578\times0.537\space atm\approx0.310\space atm\)

• Partial pressure of \(N_2F_2\):

Using the formula \(P_{N_2F_2}=x_{N_2F_2}\times P_{total}\), we substitute \(x_{N_2F_2}=0.422\) and \(P_{total}=0.537\space atm\).
\(P_{N_2F_2}=0.422\times0.537\space atm\approx0.227\space atm\)

Answer:

• Dinitrogen monoxide:
• Mole fraction: \(0.578\)
• Partial pressure: \(0.310\space atm\)
• Dinitrogen difluoride:
• Mole fraction: \(0.422\)
• Partial pressure: \(0.227\space atm\)
• Total pressure in tank: \(0.537\space atm\)